04 Rectilinear Motion Problem Set
PARTICLE RECTILINEAR MOTION | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The velocity of a particle undergoing rectilinear motion is represented by the function π£ = π‘ $ + 2π‘ m/s, where t is given in seconds. If the initial position of the particle is at the origin when the motion begins, its position after 4 seconds is closest to: m
A. 51.1 B. 37.3 C. 17.4 D. 10
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PROBLEM 2: The acceleration of a particle undergoing rectilinear motion is represented by the function π =
π m/s $ , where s is given in meters. If the particle starts from rest at the
origin, its velocity when it is 16 meters down its path of motion is closest to: m/s A. 12.1 B. 4 C. 9.25 D. 15.7
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PROBLEM 3: The acceleration of a particle undergoing rectilinear motion is represented by the function π =
π m/s $ , where s is given in meters. If the particle starts from rest at the
origin, the time it takes it to reach a point 16 meters down its path of motion is closest to: s A. 12.1 B. 4 C. 16.2 D. 6.93
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PROBLEM 4: A ball is thrown straight up in to the air from a point that is 1.2 meters above the ground. If after 3 seconds, the ball hits the ground, the maximum height that it reached is closest to: m A. 2 B. 10 C. 11.6 D. 14.2
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PROBLEM 5: An object is shot straight up in to the air from ground level with an initial velocity of 20 ft/s. The maximum height that this object will reach is closest to ft : A. 6.21 B. 9.81 C. 11.6 D. 20.4
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PROBLEM 6: An object is shot straight up in to the air from ground level with an initial velocity of 20 ft/s. The time it takes for the object to reach its maximum height is closest to s : A. . 621 B. 2.04 C. 5.1 D. 9.81
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PROBLEM 7: If a car starts from rest and travels at a constant acceleration of 3000 mi/h$ , the time is takes it to reach a distance of 1 mile is closest to: min : A. . 026 B. 93 C. 1.55 D. . 007
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PROBLEM 8: If a car starts from rest and travels at a constant acceleration of 3000 mi/h$ , its velocity at the point it reaches a distance of 1 mile from its starting point is closest to: mph : A. 1.3 B. 46.5 C. 77.5 D. 465
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PROBLEM 9: If an object initiates travel from rest at the origin, along a straight path, with an acceleration that is defined by the function: .4π‘ 0 β€ π‘ < 10 π π = 2.4 10 β€ π‘ < 20 π m/s $ 0 20 β€ π‘ < 30 π The velocity of the object after 3o seconds of motion is closest to: m/s : A. 4 B. 20 C. 24 D. 44
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PARTICLE RECTILINEAR MOTION | SOLUTIONS SOLUTION 1: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RECTILINEAR MOTION deals with the analysis of PARTICLES undergoing STRAIGHT-LINE MOTION. With STRAIGHT-LINE MOTION, we are dealing with ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
In this problem, we are given: π£ = π‘ $ + 2π‘ m/s π ? = 0 π‘? = 0
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And are asked to determine what the POSITION of the PARTICLE will be after 4 seconds in MOTION: π‘$ = 4 s Having the VELOCITY of this PARTICLE defined as a FUNCTION OF TIME, and knowing that it is undergoing CONTINUOUS RECTILINEAR MOTION, allows us to develop expressions revolving around the DISPLACEMENT and ACCELERATION of this PARTICLE over the period in which it is in MOTION. Working to QUANTIFY the DISPLACEMENT will be the center of our focus in this problem. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION. Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
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This can be more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. With a function already defined for the VELOCITY of this PARTICLE, we can start with that and work our way back to defining an expression for the DISPLACEMENT of this same PARTICLE. Starting with the expressions:
π£=
ππ ππ‘
And: π£ = π‘ $ + 2π‘ We combine them to get:
π‘ $ + 2π‘ =
ππ ππ‘
Doing a little bit of rearranging, we have: π‘ $ + 2π‘ ππ‘ = ππ
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Integrating the left side:
π‘ $ + 2π‘ ππ‘ = ππ
We define an expression for the DISPLACMENT as: π‘E π = + π‘$ 3 We know from what is given to us in the problem statement that: π‘? = 0 π ? = 0 Therefore, determining the DISPLACEMENT after 4 SECONDS, we get: (4)E π (4) = + (4)$ 3 Or: π 4 = 37.3 m It is IMPORTANT to NOTE that this is the DISPLACEMENT of the PARTICLE 4 SECONDS in to its MOTIONβ¦which may or may not coincide with its POSITION.
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However, the problem statement gives us those INITIAL CONDITIONS stating that the PARTICLE indeed starts its MOTION from the ORIGIN, which allows us to conclude that the POSITION of the PARTICLE after 4 SECONDS will be 37.3 meters. The correct answer choice is B. ππ. π π¦
SOLUTION 2: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RECTILINEAR MOTION deals with the analysis of PARTICLES undergoing STRAIGHT-LINE MOTION. With STRAIGHT-LINE MOTION, we are dealing with ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
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In this problem, we are given: π=
π m/s $
π ? = 0 π£? = 0 π‘? = 0 And are asked to determine what the VELOCITY of the PARTICLE will be when it is 16 meters away from the point motion was initiated: π $ = 16 m Having the ACCELERATION of this PARTICLE defined as a FUNCTION, and knowing that it is undergoing CONTINUOUS RECTILINEAR MOTION, allows us to develop expressions revolving around the DISPLACEMENT and VELOCITY of this PARTICLE over the period in which it is in MOTION. Working to QUANTIFY the VELOCITY will be at the center of our focus in this problem. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION.
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Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
ππ ππ‘
This is more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. For RECTILINEAR ACCELERATION, if at π‘? a PARTICLE has a specified VELOCITY as π£? , and at π‘$ , this same particle has a VELOCITY of π£$ , then the AVERAGE ACCELERATION of the PARTICLE can be expressed as:
π@AB =
π£$ β π£? π‘$ β π‘?
As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average acceleration becomes the INSTANTANEOUS ACCELERATION, such that:
π=
ππ£ ππ‘ Made with
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This is also referred to as the TIME DERIVATIVE of VELOCITY. In this problem we have a function already defined for the ACCELERATION of this PARTICLE, however, it is not given to us as a FUNCTION of TIME, but rather, one as a FUNCTION of DISPLACEMENT. This may seem like an issue, but lets recall that we have a THIRD KINEMATIC equation that we can work with. Having defined the two BASIC KINEMATIC equations representing a PARTICLE undergoing RECTILINEAR MOTION as: ππ ππ‘ ππ£ π= ππ‘ π£=
We can rearrange them, such that: ππ π£ ππ£ ππ‘ = π ππ‘ =
Set them equal to one another to cancel out the ππ‘: ππ ππ£ = π£ π
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And rearrange it one more time to derive the THIRD PARTICLE RECTILINEAR MOTION equation, which is written as: π ππ = π£ ππ£ This expression relates the DISPLACEMENT, VELOCITY, and the ACCELERATION without regard to TIMEβ¦which is what we need to solve this problem. Again, we have the ACCELERATION of this PARTICLE defined as a FUNCTION of its POSITION such that: π=
π m/s $
Plugging this in to the GENERAL EXPRESSION we just derived: π ππ = π£ ππ£ Integrating both sides:
π ππ =
π£ ππ£
We get:
π£$ β π£? =
2 E/$ π 3
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We are told that: π ? = 0 π£? = 0 π‘? = 0 π $ = 16 m With these values, we can write the quantifiable expression:
π£$ β 0 =
2 (16)E/$ 3
Or: π£$ = 9.25 m/s The correct answer choice is C. π. ππ π¦/π¬
SOLUTION 3: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RECTILINEAR MOTION deals with the analysis of PARTICLES undergoing STRAIGHT-LINE MOTION.
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With STRAIGHT-LINE MOTION, we are dealing with ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
In this problem, we are given: π=
π m/s $
π ? = 0 π£? = 0 π‘? = 0 And are asked to determine what the TIME in which it will take the PARTICLE to reach a point that is 16 meters away from the ORIGIN: π $ = 16 m Having the ACCELERATION of this PARTICLE defined as a FUNCTION, and knowing that it is undergoing CONTINUOUS RECTILINEAR MOTION, allows us to develop expressions revolving around the DISPLACEMENT and VELOCITY of this PARTICLE over the period in which it is in MOTION.
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We will focus on QUANTIFYING the TIME it will take for it to undergo a DISPLACEMENT of 16 meters. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION. Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
ππ ππ‘
This can be more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. For RECTILINEAR ACCELERATION, if at π‘? a PARTICLE has a specified VELOCITY as π£? , and at π‘$ , this same particle has a VELOCITY of π£$ , then the AVERAGE ACCELERATION of the PARTICLE can be expressed as:
π@AB =
π£$ β π£? π‘$ β π‘? Made with
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As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average acceleration becomes the INSTANTANEOUS ACCELERATION, such that:
π=
ππ£ ππ‘
This is also referred to as the TIME DERIVATIVE of VELOCITY. In this problem we have a function defined for the ACCELERATION of this PARTICLE, however, it is not given to us as a FUNCTION of TIME, but rather, one as a FUNCTION of DISPLACEMENT. We have to determine another angle to take in defining the value we are looking for. Having defined the two BASIC KINEMATIC equations representing a PARTICLE undergoing RECTILINEAR MOTION as: ππ ππ‘ ππ£ π= ππ‘ π£=
We can rearrange them, such that: ππ π£ ππ£ ππ‘ = π ππ‘ =
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Set them equal to one another to cancel out the ππ‘: ππ ππ£ = π£ π And rearrange it one more time to derive the THIRD PARTICLE RECTILINEAR MOTION equation, which is stated as: π ππ = π£ ππ£ This expression relates the DISPLACEMENT, VELOCITY, and the ACCELERATION without regard to TIMEβ¦which is what we need to get on our way towards the solution of this problem. Recall that we have the ACCELERATION of this PARTICLE defined as a FUNCTION of its POSITION such that: π=
π m/s $
Plugging this in to the GENERAL EXPRESSION we just derived: π ππ = π£ ππ£ Integrating both sides:
π ππ =
π£ ππ£ Made with
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We get:
π£$ β π£? =
2 E/$ π 3
Knowing that our PARTICLE starts from rest at the origin, we can write our VELOCITY FUNCTION as:
π£=
2 E/$ π 3
Recall that the INSTANTANEOUS VELOCITY can generally be expressed as:
π£=
ππ ππ‘
Plugging in our newly derived formula for VELOCITY, we get: 2 E/$ ππ π = 3 ππ‘ Rearranging a bit, we have:
ππ‘ =
ππ 2 E/$ π 3
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Integrating both sides, we get: O
ππ‘ = P
ππ 2 E/$ π 3
Or: π‘ = 2 3π ?/Q Which means that at: π $ = 16 We have: π‘ = 2 3(16)?/Q Or: π‘ = 6.93 s It will take the PARTICLE 6.93 SECONDS to reach a point that is 16 meters away from the ORIGIN. The correct answer choice is D. π. ππ π¬
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SOLUTION 4: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement says nothing about CONSTANT ACCELERATION, however, it can be assumed since the ball is being thrown, free from any other external forces except for the action of GRAVITY, which is a CONSTANT ACCELERATION acting towards the direction of the earth.
With ACCELERATION CONSTANT, along with the THREE BASIC KINEMATIC EXPRESSIONS, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S
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These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas will be necessary, which we will see as we make our way through the solution of this problem. So letβs define the VARIOUS STATES of MOTION in which our ball is undergoing, identifying the DATA we have, and that in which we must define. At the INSTANT the ball is PROJECTED STRAIGHT UP in to the air, we are given: π S = 1.2 m π‘S = 0 π£S = UNKNOWN At the INSTANT the ball reaches its PEAK HEIGHT in the air, we have: π£? = 0 π ? = UNKNOWN
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π‘? = UNKNOWN At the INSTANT the ball hits the GROUND: π $ = β1.2 m π‘$ = 3 s π£$ = 0 And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 9.81 m/s $ What we will do is work between STATES, defining any UNKNOWNS that we can, and moving through the solution systematically. Starting with our INITIAL and FINAL STATE, letβs QUANTIFY the INITIAL VELOCITY, π£S . Looking at the SPECIAL CASE FORMULAS we have defined for us, we will use: 1 π = π S + π£S π‘ + πS π‘ $ 2 Between these TWO STATES we have: π S = 1.2 m
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π $ = β1.2 m π‘S = 0 π‘$ = 3 s π£S = UNKNOWN π£$ = 0 πS = 9.81 m/s $ Plugging these values in, we get: 1 β1.2 m = 1.2 m + π£S (3 π ) + (β9.81 m/s $ )(3 π )$ 2 Solving for the INITIAL VELOCITY, we get: π£S = 14.3 m/s Letβs move now between the PEAK STATE and the INITIAL STATE. Looking at the SPECIAL CASE FORMULAS we have defined for us, we will use: π£ $ = π£S $ + 2 πS π β π S Between these TWO STATES we have: π P = 1.2 m π ? = UNKNOWN π‘P = 0
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π‘? = UNKNOWN π£S = 14.3 m/s π£? = 0 πS = 9.81 m/s $ Plugging these values in, we get: (0)$ = (14.3 m/s)$ + 2 (β9.81 m/s $ ) π β 1.2 m Solving for the PEAK HEIGHT we get: π ? = 11.6 m Therefore, based on the INITIAL CONDITIONS that we are presented, the ball will reach a MAXIMUM HEIGHT of 11.6 meters above the GROUND. The correct answer choice is C. ππ. π π¦
SOLUTION 5: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement says nothing about CONSTANT ACCELERATION, however, it can be assumed since the object is being shot directly upward, free from any other external forces except for the action of GRAVITY, which is a CONSTANT ACCELERATION acting towards the direction of the earth.
With ACCELERATION CONSTANT, along with the THREE BASIC KINEMATIC EXPRESSIONS, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME.
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To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas will be necessary, which we will see as we make our way through the solution of this problem. So letβs define the VARIOUS STATES of MOTION in which the object is undergoing, identifying the DATA we have, and that in which we must define. At the INSTANT the object is SHOT STRAIGHT UP in to the air, we are given: π S = 0 π‘S = 0 π£S = 20 ft/s At the INSTANT the object reaches its PEAK HEIGHT in the air, we have: π£? = 0 π ? = UNKNOWN π‘? = UNKNOWN And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 32.2 ft/s $ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the MAXIMUM HEIGHT of the OBJECT.
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Letβs QUANTIFY what we have defined between our INITIAL and PEAK HEIGHT STATE: π S = 0 π ? = UNKNOWN π‘S = 0 π‘$ = UNKNOWN π£S = 20 ft/s π£$ = 0 πS = 32.2 ft/s $ Looking at the SPECIAL CASE FORMULAS we have defined for us, we can use: π£ $ = π£S $ + 2 πS π β π S Plugging in the values we have defined, we get: (0)$ = (20 ft/s)$ + 2 (β32.2 ft/s $ ) π β 0 Solving for the PEAK HEIGHT we get: π ? = 6.21 ft Therefore, based on the INITIAL CONDITIONS that we are presented, the object will reach a MAXIMUM HEIGHT of 6.21 feet above the GROUND. The correct answer choice is A. π. ππ ππ
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SOLUTION 6: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement says nothing about CONSTANT ACCELERATION, however, it can be assumed since the object is being shot directly upward, free from any other external forces except for the action of GRAVITY, which is a CONSTANT ACCELERATION acting towards the direction of the earth.
With ACCELERATION CONSTANT, along with the three basic kinematic expressions, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S
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These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas may be necessary. So letβs define the VARIOUS STATES of MOTION in which the object is undergoing, identifying the DATA we have, and that in which we must define. At the INSTANT the object is SHOT STRAIGHT UP in to the air, we are given: π S = 0 π‘S = 0 π£S = 20 ft/s At the INSTANT the object reaches its PEAK HEIGHT in the air, we have: π£? = 0 π ? = UNKNOWN π‘? = UNKNOWN
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And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 32.2 ft/s $ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the TIME it takes for the OBJECT to reach its MAXIMUM HEIGHT. Letβs QUANTIFY what we have defined between our INITIAL and PEAK HEIGHT STATE: π S = 0 π ? = UNKNOWN π‘S = 0 π‘$ = UNKNOWN π£S = 20 ft/s π£$ = 0 πS = 32.2 ft/s $ We have a lot of good information already defined for us and looking at the SPECIAL CASE FORMULAS, we will use: π£ = π£S + πS π‘
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To define the TIME it takes to reach the PEAK of the MOTION, lets plug in the values we have defined: 0 = 20 ft/s + (β32.2 ft/s $ )π‘ Solving for the TIME, we get: π‘? = 6.21 s Therefore, based on the INITIAL CONDITIONS that we are presented, the object will reach its MAXIMUM HEIGHT .621 seconds after it is launched. The correct answer choice is A. . πππ π¬
SOLUTION 7: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement states that a car is undergoing RECTILINEAR MOTION with a CONSTANT ACCELERATION as defined.
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With ACCELERATION CONSTANT, along with the three basic kinematic expressions, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas may be necessary. So letβs define the VARIOUS STATES of MOTION in which the car is undergoing, identifying the DATA we have, and that in which we must define.
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At the INSTANT the car starts its STRAIGHT LINE MOTION, we are given: π S = 0 π‘S = 0 π£S = 0 mph At the INSTANT the car reaches a point 1 mile down the road, we have: π£? = UNKNOWN π ? = 1 mile π‘? = UNKNOWN And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 3000 mi/h$ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the TIME it takes for the CAR to reach the desired point of analysis. Letβs QUANTIFY what we have defined between our INITIAL and 1 MILE DISTANCE STATE: π S = 0 π ? = 1 mile π‘S = 0
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π‘$ = UNKNOWN π£S = 0 ft/s π£$ = UNKNOWN πS = 3000 mi/h$ We have a lot of good information already defined for us and looking at the SPECIAL CASE FORMULAS, we will use: 1 π = π S + π£S π‘ + πS π‘ $ 2 This formula will allow us to define the TIME it takes to reach a point that is 1 Mile down the path of travel. Plugging these values in, we get: 1 1 mile = 0 + (0)π‘ + (3000 mi/h$ )π‘ $ 2 Isolating an solving for the TIME: π‘? = .026 hr This is the correct TIME it will take for the car to reach a point that is 1 mile down the path of motion, however, the problem statement is asking for the UNITS to be in MINUTES, so we must convert.
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Doing so, we get:
. 026 hr
60 min 1 hr
Or: π‘? = 1.55 min Therefore, based on the INITIAL CONDITIONS that we are presented, the car will reach a point that is 1 mile down the road in 1.55 minutes. The correct answer choice is C. π. ππ π¦π’π§
SOLUTION 8: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement states that a car is undergoing RECTILINEAR MOTION with a CONSTANT ACCELERATION as defined.
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With ACCELERATION CONSTANT, along with the THREE BASIC KINEMATIC EXPRESSIONS, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas may be necessary. So letβs define the VARIOUS STATES of MOTION in which the car is undergoing, identifying the DATA we have, and that in which we must define.
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At the INSTANT the car starts its STRAIGHT LINE MOTION, we are given: π S = 0 π‘S = 0 π£S = 0 mph At the INSTANT the object car reaches a point 1 mile down the road, we have: π£? = UNKNOWN π ? = 1 mile π‘? = UNKNOWN And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 3000 mi/h$ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the TIME it takes for the CAR to reach the point as defined. Letβs lay out what we have defined between our INITIAL and 1 MILE DISTANCE STATE: π S = 0 π ? = 1 mile π‘S = 0
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π‘? = UNKNOWN π£S = 0 ft/s π£$ = UNKNOWN πS = 3000 mi/h$ We have a lot of good information already defined for us and looking at the SPECIAL CASE FORMULAS, we will use: 1 π = π S + π£S π‘ + πS π‘ $ 2 To define the TIME it takes to reach a point that is 1 Mile down the path of travel. Plugging our values in, we have: 1 1 mile = 0 + (0)π‘ + (3000 mi/h$ )π‘ $ 2 Solving for the TIME, we get: π‘? = .026 hr We can now use the SPECIAL CASE formula: π£ = π£S + πS π‘ To QUANTIFY the VELOCITY 1 MILE down the road.
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Plugging in our DATA, we have: π£ = (0) + (3000 mi/h$ )(.026 h) Or: π£ = 77.5 mph Therefore, based on the INITIAL CONDITIONS that we are presented, the car will be going 77.5 mph at a point that is 1 mile down the road from where it initiated its motion. The correct answer choice is C. ππ. π π¦π©π‘
SOLUTION 9: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When observing a PARTICLE in MOTION within an established FRAME OF REFERENCE, COORDINATE SYSTEM, or INTERVAL, we can say that it is undergoing RECTILINEAR MOTION if its POSITION VECTOR at POINT A is PARALLEL with its POSITION VECTOR at POINT B.
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This creates a scenario of ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
The MOTION of a PARTICLE can be expressed as a FUNCTION OF TIME, in which, will identify whether the PARTICLE is undergoing CONTINUOUS or ERRATIC RECTILINEAR MOTION. When discussing CONTINUOUS RECTILINEAR MOTION, the function representing the POSITION of the PARTICLE is composed of a SINGLE FUNCTION. This is not the case when discussing ERRATIC RECTILINEAR MOTION, wherein which, the FUNCTION will be defined by multiple FUNCTIONS over different PERIODS in the MOTION of the PARTICLE. In this problem, we are given the ACCELERATION of the PARTICLE as a PIECEWISE FUNCTION, defined as: . 4π‘ 0 β€ π‘ < 10 π π = 2.4 10 β€ π‘ < 20 π m/s $ 0 20 β€ π‘ < 30 π
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When dealing with ERRATIC RECTILINEAR MOTION, the POSITION versus TIME DEPENDENCE can not be fully captured by a single function over the PERIOD of analysis. We need to take in to account multiple equations that relate to different TIME PERIODS covering the overall MOTION of the PARTICLE. However, as with CONTIUOUS RECTILINEAR MOTION, the general DISPLACEMENT, VELOCITY, and ACCELERATION functions are still valid, we just have to be cognizant on where each period begins and ends and ensure we are applying the proper FUNCTION that defines the INTERVAL, or DOMAIN, we are in. So with that, letβs work our solution. We are asked to determine the VELOCITY of this PARTICLE 30 seconds in to its MOTION. Taking a look at the PIECEWISE FUNCTION that defines the MOTION of this PARTICLE, we see that this analysis will span all three INTERVALS. We will approach this as we would any other problem, taking it one INTERVAL at a time until the full solution has been derived. Having the ACCELERATION of this PARTICLE defined as a FUNCTION, and knowing that it is undergoing ERRATIC RECTILINEAR MOTION, which can be treated as CONTINUOUS RECTILINEAR MOTION within each INTERVAL, allows us to develop
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expressions revolving around the DISPLACEMENT and VELOCITY of the PARTICLE over the period in which it is in MOTION. We will focus on QUANTIFYING the VELOCITY in this problem. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION. Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
ππ ππ‘
This is more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. However, DISPLACEMENT is not something we are given right off the top, so we have to see how we can work this problem from a different angle. Letβs recap RECTILINEAR ACCELERATION.
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If at π‘? a PARTICLE has a specified VELOCITY as π£? , and at π‘$ , this same particle has a VELOCITY of π£$ , then the AVERAGE ACCELERATION of the PARTICLE can be expressed as:
π@AB =
π£$ β π£? π‘$ β π‘?
As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average acceleration becomes the INSTANTANEOUS ACCELERATION, such that:
π=
ππ£ ππ‘
This is also referred to as the TIME DERIVATIVE of VELOCITY. Having the ACCELERATION defined in each INTERVAL, we can plug in each expression and derive an expression for the VELOCITY in the sane interval. Letβs do that starting with the FIRST INTERVAL, which gives us:
. 4π‘ =
ππ£ ππ‘
Rearranging a bit: . 4π‘ππ‘ = ππ£
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Which gives us the VELOCITY FUNCTION: π£(π‘) = .2π‘ $ For the SECOND INTERVAL, we have:
2.4 =
ππ£ ππ‘
Rearranging a bit: 2.4ππ‘ = ππ£ Which gives us the VELOCITY FUNCTION: π£(π‘) = 2.4π‘ Lastly, for the THIRD INTERVAL, we have:
0=
ππ£ ππ‘
Which means that with NO ACCELERATION, the VELOCITY will not CHANGEβ¦so no analysis needs to be done in this INTERVAL, which simplifies the problem a bit.
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We now have: .2π‘ $ 0 β€ π‘ < 10 π π£ = 2.4π‘ 10 β€ π‘ < 20 π m/s 0 20 β€ π‘ < 30 π The VELOCITY after 30 SECONDS in MOTION will then be: π£ = π£ 0 β€ π‘ < 10 π + π£ 10 β€ π‘ < 20 π + 0 Or: π£ = .2(10)$ + 2.4(10) + 0 Which gives us: π£ = 44 m/s Therefore, based on the CHARACTERISTICS given for this PARTICLE, it will be going 44 m/s after 30 seconds in motion. The correct answer choice is D. ππ π¦/π¬
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PROBLEM 1: The velocity of a particle undergoing rectilinear motion is represented by the function π£ = π‘ $ + 2π‘ m/s, where t is given in seconds. If the initial position of the particle is at the origin when the motion begins, its position after 4 seconds is closest to: m
A. 51.1 B. 37.3 C. 17.4 D. 10
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PROBLEM 2: The acceleration of a particle undergoing rectilinear motion is represented by the function π =
π m/s $ , where s is given in meters. If the particle starts from rest at the
origin, its velocity when it is 16 meters down its path of motion is closest to: m/s A. 12.1 B. 4 C. 9.25 D. 15.7
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PROBLEM 3: The acceleration of a particle undergoing rectilinear motion is represented by the function π =
π m/s $ , where s is given in meters. If the particle starts from rest at the
origin, the time it takes it to reach a point 16 meters down its path of motion is closest to: s A. 12.1 B. 4 C. 16.2 D. 6.93
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PROBLEM 4: A ball is thrown straight up in to the air from a point that is 1.2 meters above the ground. If after 3 seconds, the ball hits the ground, the maximum height that it reached is closest to: m A. 2 B. 10 C. 11.6 D. 14.2
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PROBLEM 5: An object is shot straight up in to the air from ground level with an initial velocity of 20 ft/s. The maximum height that this object will reach is closest to ft : A. 6.21 B. 9.81 C. 11.6 D. 20.4
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PROBLEM 6: An object is shot straight up in to the air from ground level with an initial velocity of 20 ft/s. The time it takes for the object to reach its maximum height is closest to s : A. . 621 B. 2.04 C. 5.1 D. 9.81
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PROBLEM 7: If a car starts from rest and travels at a constant acceleration of 3000 mi/h$ , the time is takes it to reach a distance of 1 mile is closest to: min : A. . 026 B. 93 C. 1.55 D. . 007
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PROBLEM 8: If a car starts from rest and travels at a constant acceleration of 3000 mi/h$ , its velocity at the point it reaches a distance of 1 mile from its starting point is closest to: mph : A. 1.3 B. 46.5 C. 77.5 D. 465
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PROBLEM 9: If an object initiates travel from rest at the origin, along a straight path, with an acceleration that is defined by the function: .4π‘ 0 β€ π‘ < 10 π π = 2.4 10 β€ π‘ < 20 π m/s $ 0 20 β€ π‘ < 30 π The velocity of the object after 3o seconds of motion is closest to: m/s : A. 4 B. 20 C. 24 D. 44
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PARTICLE RECTILINEAR MOTION | SOLUTIONS SOLUTION 1: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RECTILINEAR MOTION deals with the analysis of PARTICLES undergoing STRAIGHT-LINE MOTION. With STRAIGHT-LINE MOTION, we are dealing with ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
In this problem, we are given: π£ = π‘ $ + 2π‘ m/s π ? = 0 π‘? = 0
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And are asked to determine what the POSITION of the PARTICLE will be after 4 seconds in MOTION: π‘$ = 4 s Having the VELOCITY of this PARTICLE defined as a FUNCTION OF TIME, and knowing that it is undergoing CONTINUOUS RECTILINEAR MOTION, allows us to develop expressions revolving around the DISPLACEMENT and ACCELERATION of this PARTICLE over the period in which it is in MOTION. Working to QUANTIFY the DISPLACEMENT will be the center of our focus in this problem. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION. Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
ππ ππ‘ Made with
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This can be more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. With a function already defined for the VELOCITY of this PARTICLE, we can start with that and work our way back to defining an expression for the DISPLACEMENT of this same PARTICLE. Starting with the expressions:
π£=
ππ ππ‘
And: π£ = π‘ $ + 2π‘ We combine them to get:
π‘ $ + 2π‘ =
ππ ππ‘
Doing a little bit of rearranging, we have: π‘ $ + 2π‘ ππ‘ = ππ
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Integrating the left side:
π‘ $ + 2π‘ ππ‘ = ππ
We define an expression for the DISPLACMENT as: π‘E π = + π‘$ 3 We know from what is given to us in the problem statement that: π‘? = 0 π ? = 0 Therefore, determining the DISPLACEMENT after 4 SECONDS, we get: (4)E π (4) = + (4)$ 3 Or: π 4 = 37.3 m It is IMPORTANT to NOTE that this is the DISPLACEMENT of the PARTICLE 4 SECONDS in to its MOTIONβ¦which may or may not coincide with its POSITION.
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However, the problem statement gives us those INITIAL CONDITIONS stating that the PARTICLE indeed starts its MOTION from the ORIGIN, which allows us to conclude that the POSITION of the PARTICLE after 4 SECONDS will be 37.3 meters. The correct answer choice is B. ππ. π π¦
SOLUTION 2: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RECTILINEAR MOTION deals with the analysis of PARTICLES undergoing STRAIGHT-LINE MOTION. With STRAIGHT-LINE MOTION, we are dealing with ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
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In this problem, we are given: π=
π m/s $
π ? = 0 π£? = 0 π‘? = 0 And are asked to determine what the VELOCITY of the PARTICLE will be when it is 16 meters away from the point motion was initiated: π $ = 16 m Having the ACCELERATION of this PARTICLE defined as a FUNCTION, and knowing that it is undergoing CONTINUOUS RECTILINEAR MOTION, allows us to develop expressions revolving around the DISPLACEMENT and VELOCITY of this PARTICLE over the period in which it is in MOTION. Working to QUANTIFY the VELOCITY will be at the center of our focus in this problem. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION.
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Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
ππ ππ‘
This is more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. For RECTILINEAR ACCELERATION, if at π‘? a PARTICLE has a specified VELOCITY as π£? , and at π‘$ , this same particle has a VELOCITY of π£$ , then the AVERAGE ACCELERATION of the PARTICLE can be expressed as:
π@AB =
π£$ β π£? π‘$ β π‘?
As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average acceleration becomes the INSTANTANEOUS ACCELERATION, such that:
π=
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This is also referred to as the TIME DERIVATIVE of VELOCITY. In this problem we have a function already defined for the ACCELERATION of this PARTICLE, however, it is not given to us as a FUNCTION of TIME, but rather, one as a FUNCTION of DISPLACEMENT. This may seem like an issue, but lets recall that we have a THIRD KINEMATIC equation that we can work with. Having defined the two BASIC KINEMATIC equations representing a PARTICLE undergoing RECTILINEAR MOTION as: ππ ππ‘ ππ£ π= ππ‘ π£=
We can rearrange them, such that: ππ π£ ππ£ ππ‘ = π ππ‘ =
Set them equal to one another to cancel out the ππ‘: ππ ππ£ = π£ π
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And rearrange it one more time to derive the THIRD PARTICLE RECTILINEAR MOTION equation, which is written as: π ππ = π£ ππ£ This expression relates the DISPLACEMENT, VELOCITY, and the ACCELERATION without regard to TIMEβ¦which is what we need to solve this problem. Again, we have the ACCELERATION of this PARTICLE defined as a FUNCTION of its POSITION such that: π=
π m/s $
Plugging this in to the GENERAL EXPRESSION we just derived: π ππ = π£ ππ£ Integrating both sides:
π ππ =
π£ ππ£
We get:
π£$ β π£? =
2 E/$ π 3
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We are told that: π ? = 0 π£? = 0 π‘? = 0 π $ = 16 m With these values, we can write the quantifiable expression:
π£$ β 0 =
2 (16)E/$ 3
Or: π£$ = 9.25 m/s The correct answer choice is C. π. ππ π¦/π¬
SOLUTION 3: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RECTILINEAR MOTION deals with the analysis of PARTICLES undergoing STRAIGHT-LINE MOTION.
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With STRAIGHT-LINE MOTION, we are dealing with ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
In this problem, we are given: π=
π m/s $
π ? = 0 π£? = 0 π‘? = 0 And are asked to determine what the TIME in which it will take the PARTICLE to reach a point that is 16 meters away from the ORIGIN: π $ = 16 m Having the ACCELERATION of this PARTICLE defined as a FUNCTION, and knowing that it is undergoing CONTINUOUS RECTILINEAR MOTION, allows us to develop expressions revolving around the DISPLACEMENT and VELOCITY of this PARTICLE over the period in which it is in MOTION.
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We will focus on QUANTIFYING the TIME it will take for it to undergo a DISPLACEMENT of 16 meters. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION. Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
ππ ππ‘
This can be more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. For RECTILINEAR ACCELERATION, if at π‘? a PARTICLE has a specified VELOCITY as π£? , and at π‘$ , this same particle has a VELOCITY of π£$ , then the AVERAGE ACCELERATION of the PARTICLE can be expressed as:
π@AB =
π£$ β π£? π‘$ β π‘? Made with
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As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average acceleration becomes the INSTANTANEOUS ACCELERATION, such that:
π=
ππ£ ππ‘
This is also referred to as the TIME DERIVATIVE of VELOCITY. In this problem we have a function defined for the ACCELERATION of this PARTICLE, however, it is not given to us as a FUNCTION of TIME, but rather, one as a FUNCTION of DISPLACEMENT. We have to determine another angle to take in defining the value we are looking for. Having defined the two BASIC KINEMATIC equations representing a PARTICLE undergoing RECTILINEAR MOTION as: ππ ππ‘ ππ£ π= ππ‘ π£=
We can rearrange them, such that: ππ π£ ππ£ ππ‘ = π ππ‘ =
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Set them equal to one another to cancel out the ππ‘: ππ ππ£ = π£ π And rearrange it one more time to derive the THIRD PARTICLE RECTILINEAR MOTION equation, which is stated as: π ππ = π£ ππ£ This expression relates the DISPLACEMENT, VELOCITY, and the ACCELERATION without regard to TIMEβ¦which is what we need to get on our way towards the solution of this problem. Recall that we have the ACCELERATION of this PARTICLE defined as a FUNCTION of its POSITION such that: π=
π m/s $
Plugging this in to the GENERAL EXPRESSION we just derived: π ππ = π£ ππ£ Integrating both sides:
π ππ =
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We get:
π£$ β π£? =
2 E/$ π 3
Knowing that our PARTICLE starts from rest at the origin, we can write our VELOCITY FUNCTION as:
π£=
2 E/$ π 3
Recall that the INSTANTANEOUS VELOCITY can generally be expressed as:
π£=
ππ ππ‘
Plugging in our newly derived formula for VELOCITY, we get: 2 E/$ ππ π = 3 ππ‘ Rearranging a bit, we have:
ππ‘ =
ππ 2 E/$ π 3
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Integrating both sides, we get: O
ππ‘ = P
ππ 2 E/$ π 3
Or: π‘ = 2 3π ?/Q Which means that at: π $ = 16 We have: π‘ = 2 3(16)?/Q Or: π‘ = 6.93 s It will take the PARTICLE 6.93 SECONDS to reach a point that is 16 meters away from the ORIGIN. The correct answer choice is D. π. ππ π¬
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SOLUTION 4: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement says nothing about CONSTANT ACCELERATION, however, it can be assumed since the ball is being thrown, free from any other external forces except for the action of GRAVITY, which is a CONSTANT ACCELERATION acting towards the direction of the earth.
With ACCELERATION CONSTANT, along with the THREE BASIC KINEMATIC EXPRESSIONS, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S
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These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas will be necessary, which we will see as we make our way through the solution of this problem. So letβs define the VARIOUS STATES of MOTION in which our ball is undergoing, identifying the DATA we have, and that in which we must define. At the INSTANT the ball is PROJECTED STRAIGHT UP in to the air, we are given: π S = 1.2 m π‘S = 0 π£S = UNKNOWN At the INSTANT the ball reaches its PEAK HEIGHT in the air, we have: π£? = 0 π ? = UNKNOWN
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π‘? = UNKNOWN At the INSTANT the ball hits the GROUND: π $ = β1.2 m π‘$ = 3 s π£$ = 0 And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 9.81 m/s $ What we will do is work between STATES, defining any UNKNOWNS that we can, and moving through the solution systematically. Starting with our INITIAL and FINAL STATE, letβs QUANTIFY the INITIAL VELOCITY, π£S . Looking at the SPECIAL CASE FORMULAS we have defined for us, we will use: 1 π = π S + π£S π‘ + πS π‘ $ 2 Between these TWO STATES we have: π S = 1.2 m
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π $ = β1.2 m π‘S = 0 π‘$ = 3 s π£S = UNKNOWN π£$ = 0 πS = 9.81 m/s $ Plugging these values in, we get: 1 β1.2 m = 1.2 m + π£S (3 π ) + (β9.81 m/s $ )(3 π )$ 2 Solving for the INITIAL VELOCITY, we get: π£S = 14.3 m/s Letβs move now between the PEAK STATE and the INITIAL STATE. Looking at the SPECIAL CASE FORMULAS we have defined for us, we will use: π£ $ = π£S $ + 2 πS π β π S Between these TWO STATES we have: π P = 1.2 m π ? = UNKNOWN π‘P = 0
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π‘? = UNKNOWN π£S = 14.3 m/s π£? = 0 πS = 9.81 m/s $ Plugging these values in, we get: (0)$ = (14.3 m/s)$ + 2 (β9.81 m/s $ ) π β 1.2 m Solving for the PEAK HEIGHT we get: π ? = 11.6 m Therefore, based on the INITIAL CONDITIONS that we are presented, the ball will reach a MAXIMUM HEIGHT of 11.6 meters above the GROUND. The correct answer choice is C. ππ. π π¦
SOLUTION 5: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement says nothing about CONSTANT ACCELERATION, however, it can be assumed since the object is being shot directly upward, free from any other external forces except for the action of GRAVITY, which is a CONSTANT ACCELERATION acting towards the direction of the earth.
With ACCELERATION CONSTANT, along with the THREE BASIC KINEMATIC EXPRESSIONS, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME.
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To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas will be necessary, which we will see as we make our way through the solution of this problem. So letβs define the VARIOUS STATES of MOTION in which the object is undergoing, identifying the DATA we have, and that in which we must define. At the INSTANT the object is SHOT STRAIGHT UP in to the air, we are given: π S = 0 π‘S = 0 π£S = 20 ft/s At the INSTANT the object reaches its PEAK HEIGHT in the air, we have: π£? = 0 π ? = UNKNOWN π‘? = UNKNOWN And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 32.2 ft/s $ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the MAXIMUM HEIGHT of the OBJECT.
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Letβs QUANTIFY what we have defined between our INITIAL and PEAK HEIGHT STATE: π S = 0 π ? = UNKNOWN π‘S = 0 π‘$ = UNKNOWN π£S = 20 ft/s π£$ = 0 πS = 32.2 ft/s $ Looking at the SPECIAL CASE FORMULAS we have defined for us, we can use: π£ $ = π£S $ + 2 πS π β π S Plugging in the values we have defined, we get: (0)$ = (20 ft/s)$ + 2 (β32.2 ft/s $ ) π β 0 Solving for the PEAK HEIGHT we get: π ? = 6.21 ft Therefore, based on the INITIAL CONDITIONS that we are presented, the object will reach a MAXIMUM HEIGHT of 6.21 feet above the GROUND. The correct answer choice is A. π. ππ ππ
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SOLUTION 6: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement says nothing about CONSTANT ACCELERATION, however, it can be assumed since the object is being shot directly upward, free from any other external forces except for the action of GRAVITY, which is a CONSTANT ACCELERATION acting towards the direction of the earth.
With ACCELERATION CONSTANT, along with the three basic kinematic expressions, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S
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These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas may be necessary. So letβs define the VARIOUS STATES of MOTION in which the object is undergoing, identifying the DATA we have, and that in which we must define. At the INSTANT the object is SHOT STRAIGHT UP in to the air, we are given: π S = 0 π‘S = 0 π£S = 20 ft/s At the INSTANT the object reaches its PEAK HEIGHT in the air, we have: π£? = 0 π ? = UNKNOWN π‘? = UNKNOWN
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And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 32.2 ft/s $ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the TIME it takes for the OBJECT to reach its MAXIMUM HEIGHT. Letβs QUANTIFY what we have defined between our INITIAL and PEAK HEIGHT STATE: π S = 0 π ? = UNKNOWN π‘S = 0 π‘$ = UNKNOWN π£S = 20 ft/s π£$ = 0 πS = 32.2 ft/s $ We have a lot of good information already defined for us and looking at the SPECIAL CASE FORMULAS, we will use: π£ = π£S + πS π‘
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To define the TIME it takes to reach the PEAK of the MOTION, lets plug in the values we have defined: 0 = 20 ft/s + (β32.2 ft/s $ )π‘ Solving for the TIME, we get: π‘? = 6.21 s Therefore, based on the INITIAL CONDITIONS that we are presented, the object will reach its MAXIMUM HEIGHT .621 seconds after it is launched. The correct answer choice is A. . πππ π¬
SOLUTION 7: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement states that a car is undergoing RECTILINEAR MOTION with a CONSTANT ACCELERATION as defined.
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With ACCELERATION CONSTANT, along with the three basic kinematic expressions, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas may be necessary. So letβs define the VARIOUS STATES of MOTION in which the car is undergoing, identifying the DATA we have, and that in which we must define.
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At the INSTANT the car starts its STRAIGHT LINE MOTION, we are given: π S = 0 π‘S = 0 π£S = 0 mph At the INSTANT the car reaches a point 1 mile down the road, we have: π£? = UNKNOWN π ? = 1 mile π‘? = UNKNOWN And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 3000 mi/h$ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the TIME it takes for the CAR to reach the desired point of analysis. Letβs QUANTIFY what we have defined between our INITIAL and 1 MILE DISTANCE STATE: π S = 0 π ? = 1 mile π‘S = 0
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π‘$ = UNKNOWN π£S = 0 ft/s π£$ = UNKNOWN πS = 3000 mi/h$ We have a lot of good information already defined for us and looking at the SPECIAL CASE FORMULAS, we will use: 1 π = π S + π£S π‘ + πS π‘ $ 2 This formula will allow us to define the TIME it takes to reach a point that is 1 Mile down the path of travel. Plugging these values in, we get: 1 1 mile = 0 + (0)π‘ + (3000 mi/h$ )π‘ $ 2 Isolating an solving for the TIME: π‘? = .026 hr This is the correct TIME it will take for the car to reach a point that is 1 mile down the path of motion, however, the problem statement is asking for the UNITS to be in MINUTES, so we must convert.
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Doing so, we get:
. 026 hr
60 min 1 hr
Or: π‘? = 1.55 min Therefore, based on the INITIAL CONDITIONS that we are presented, the car will reach a point that is 1 mile down the road in 1.55 minutes. The correct answer choice is C. π. ππ π¦π’π§
SOLUTION 8: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A SPECIAL CASE that we will often encounter in PARTICLE RECTILINEAR MOTION is when the ACCELERATION of a PARTICLE is NOT CHANGING over timeβ¦or otherwise, when a PARTICLE is experiencing CONSTANT ACCELERATION. Our problem statement states that a car is undergoing RECTILINEAR MOTION with a CONSTANT ACCELERATION as defined.
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With ACCELERATION CONSTANT, along with the THREE BASIC KINEMATIC EXPRESSIONS, we have three additional SPECIAL CASE FORMULAS that we can use in our problems, which are stated as: π£ = π£S + πS π‘ 1 π = π S + π£S π‘ + πS π‘ $ 2 π£ $ = π£S $ + 2 πS π β π S These GENERAL FORMULAS can be found under the section PARTICLE RECTILINEAR MOTION on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For any problem involving CONSTANT ACCELERATION of a PARTICLE undergoing RECTILINEAR MOTION, it is IMPORTANT to NOTE that not one of these equations is INDEPENDENT, only two unknown variables can be defined between a set of equations between two statesβ¦unknowns being DISPLACEMENT, VELOCITY, ACCELERATION, and TIME. To fully define the characteristics of a PARTICLE, the use of a COMBINATION of these formulas may be necessary. So letβs define the VARIOUS STATES of MOTION in which the car is undergoing, identifying the DATA we have, and that in which we must define.
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At the INSTANT the car starts its STRAIGHT LINE MOTION, we are given: π S = 0 π‘S = 0 π£S = 0 mph At the INSTANT the object car reaches a point 1 mile down the road, we have: π£? = UNKNOWN π ? = 1 mile π‘? = UNKNOWN And through this whole sequence of MOTION, ACCELERATION remains CONSTANT at: πS = 3000 mi/h$ What we will do is work between the TWO STATES, defining the UNKNOWNS that we can, moving through the solution systematically in route to defining the TIME it takes for the CAR to reach the point as defined. Letβs lay out what we have defined between our INITIAL and 1 MILE DISTANCE STATE: π S = 0 π ? = 1 mile π‘S = 0
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π‘? = UNKNOWN π£S = 0 ft/s π£$ = UNKNOWN πS = 3000 mi/h$ We have a lot of good information already defined for us and looking at the SPECIAL CASE FORMULAS, we will use: 1 π = π S + π£S π‘ + πS π‘ $ 2 To define the TIME it takes to reach a point that is 1 Mile down the path of travel. Plugging our values in, we have: 1 1 mile = 0 + (0)π‘ + (3000 mi/h$ )π‘ $ 2 Solving for the TIME, we get: π‘? = .026 hr We can now use the SPECIAL CASE formula: π£ = π£S + πS π‘ To QUANTIFY the VELOCITY 1 MILE down the road.
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Plugging in our DATA, we have: π£ = (0) + (3000 mi/h$ )(.026 h) Or: π£ = 77.5 mph Therefore, based on the INITIAL CONDITIONS that we are presented, the car will be going 77.5 mph at a point that is 1 mile down the road from where it initiated its motion. The correct answer choice is C. ππ. π π¦π©π‘
SOLUTION 9: The TOPIC of PARTICLE RECTILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When observing a PARTICLE in MOTION within an established FRAME OF REFERENCE, COORDINATE SYSTEM, or INTERVAL, we can say that it is undergoing RECTILINEAR MOTION if its POSITION VECTOR at POINT A is PARALLEL with its POSITION VECTOR at POINT B.
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This creates a scenario of ONE-DIMENSIONAL movement down a DEFINED PATH, which generally can be illustrated as:
The MOTION of a PARTICLE can be expressed as a FUNCTION OF TIME, in which, will identify whether the PARTICLE is undergoing CONTINUOUS or ERRATIC RECTILINEAR MOTION. When discussing CONTINUOUS RECTILINEAR MOTION, the function representing the POSITION of the PARTICLE is composed of a SINGLE FUNCTION. This is not the case when discussing ERRATIC RECTILINEAR MOTION, wherein which, the FUNCTION will be defined by multiple FUNCTIONS over different PERIODS in the MOTION of the PARTICLE. In this problem, we are given the ACCELERATION of the PARTICLE as a PIECEWISE FUNCTION, defined as: . 4π‘ 0 β€ π‘ < 10 π π = 2.4 10 β€ π‘ < 20 π m/s $ 0 20 β€ π‘ < 30 π
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When dealing with ERRATIC RECTILINEAR MOTION, the POSITION versus TIME DEPENDENCE can not be fully captured by a single function over the PERIOD of analysis. We need to take in to account multiple equations that relate to different TIME PERIODS covering the overall MOTION of the PARTICLE. However, as with CONTIUOUS RECTILINEAR MOTION, the general DISPLACEMENT, VELOCITY, and ACCELERATION functions are still valid, we just have to be cognizant on where each period begins and ends and ensure we are applying the proper FUNCTION that defines the INTERVAL, or DOMAIN, we are in. So with that, letβs work our solution. We are asked to determine the VELOCITY of this PARTICLE 30 seconds in to its MOTION. Taking a look at the PIECEWISE FUNCTION that defines the MOTION of this PARTICLE, we see that this analysis will span all three INTERVALS. We will approach this as we would any other problem, taking it one INTERVAL at a time until the full solution has been derived. Having the ACCELERATION of this PARTICLE defined as a FUNCTION, and knowing that it is undergoing ERRATIC RECTILINEAR MOTION, which can be treated as CONTINUOUS RECTILINEAR MOTION within each INTERVAL, allows us to develop
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expressions revolving around the DISPLACEMENT and VELOCITY of the PARTICLE over the period in which it is in MOTION. We will focus on QUANTIFYING the VELOCITY in this problem. Letβs take a step back and put some context around the measurement of VELOCITY for a PARTICLE undergoing RECTILINEAR MOTION. Given a PERIOD of TIME starting at π‘? and ending at π‘$ , the AVERAGE VELOCITY of a PARTICLE can be modeled as the CHANGE IN POSITION over the CHANGE IN TIME, or in formulaic terms:
π£@AB =
π $ β π ? π‘$ β π‘?
As this CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, such that:
π£=
ππ ππ‘
This is more memorably referred to as the TIME DERIVATIVE of the DISPLACEMENT. However, DISPLACEMENT is not something we are given right off the top, so we have to see how we can work this problem from a different angle. Letβs recap RECTILINEAR ACCELERATION.
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If at π‘? a PARTICLE has a specified VELOCITY as π£? , and at π‘$ , this same particle has a VELOCITY of π£$ , then the AVERAGE ACCELERATION of the PARTICLE can be expressed as:
π@AB =
π£$ β π£? π‘$ β π‘?
As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average acceleration becomes the INSTANTANEOUS ACCELERATION, such that:
π=
ππ£ ππ‘
This is also referred to as the TIME DERIVATIVE of VELOCITY. Having the ACCELERATION defined in each INTERVAL, we can plug in each expression and derive an expression for the VELOCITY in the sane interval. Letβs do that starting with the FIRST INTERVAL, which gives us:
. 4π‘ =
ππ£ ππ‘
Rearranging a bit: . 4π‘ππ‘ = ππ£
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Which gives us the VELOCITY FUNCTION: π£(π‘) = .2π‘ $ For the SECOND INTERVAL, we have:
2.4 =
ππ£ ππ‘
Rearranging a bit: 2.4ππ‘ = ππ£ Which gives us the VELOCITY FUNCTION: π£(π‘) = 2.4π‘ Lastly, for the THIRD INTERVAL, we have:
0=
ππ£ ππ‘
Which means that with NO ACCELERATION, the VELOCITY will not CHANGEβ¦so no analysis needs to be done in this INTERVAL, which simplifies the problem a bit.
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We now have: .2π‘ $ 0 β€ π‘ < 10 π π£ = 2.4π‘ 10 β€ π‘ < 20 π m/s 0 20 β€ π‘ < 30 π The VELOCITY after 30 SECONDS in MOTION will then be: π£ = π£ 0 β€ π‘ < 10 π + π£ 10 β€ π‘ < 20 π + 0 Or: π£ = .2(10)$ + 2.4(10) + 0 Which gives us: π£ = 44 m/s Therefore, based on the CHARACTERISTICS given for this PARTICLE, it will be going 44 m/s after 30 seconds in motion. The correct answer choice is D. ππ π¦/π¬
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