09 Relative Motion Concept Overview
RELATIVE MOTION | CONCEPT OVERVIEW
The TOPIC of RELATIVE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 and 73 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: When we are analyzing OBJECTS, or PARTICLES, that are in MOTION, we inherently start with some reference frame. Whether we set this frame consciously or subconsciously, all the measurements are drawn based on that particular point of reference. Our daily lives are lived within the first person REFERENCE FRAMEβ¦this is where we by default define the characteristics of MOTION such as VELOCITY and ACCELERATION, more accurately described as RELATIVE VELOCITY and RELATIVE ACCELERATION, as we will define. These characteristic measurements of motion can seem dramatically different from another individualβs perspective because their REFERENCE FRAME isnβt set at the same point as yours.
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To put an illustration around this, say that you are standing on the beaches of Kauai gazing in to the water at a couple on paddle boarders. You have your eyes set on the person in front as they are paddling across your vision at some rate of speed. The person on the second paddle board, trailing the individual in front, is also watching the leading individual as they paddle themselves to keep up. Simple observation tells us that the motion of the individual in front paddling will look different to us standing on that stagnant shore than it does to the individual trailing behind paddling at some rate of speed as well. To the individual trailing the person in front, the motion of that leading paddle boarder looks much slower than to the eyes of you who are standing still on the shore. These observations are the result of RELATIVE MOTION. RELATIVE MOTION is the calculation of the MOTION of an OBJECT WITH REGARD TO SOME OTHER MOVING OBJECT, or stationary object for that matter. The motion is not calculated with reference to the earth, but the velocity of the object in reference to the other moving object as if it were in a static state. RELATIVE MOTION is a concept, and its calculation occurs in the context of RELATIVE VELOCITY, RELATIVE SPEED, or RELATIVE ACCELERATION.
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Letβs take this illustration and set up visual context around viewing and analyzing RELATIVE MOTION problems. Since you are standing on the beach, not moving, we can establish your point of reference as a FIXED REFERENCE FRAME, such that:
For simplification purposes, you as the initial point of reference will be considered the ORIGIN, or PARTCILE O, from this point moving forward.
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With this ORIGIN established, we will define two paddle boarders off in the distance, within a SECOND REFERENCE FRAME. The paddle boarder in the back will be PARTICLE B and the paddle boarder in the front will be PARTICLE A, as such:
To properly analyze this scenario, we must set up this SECOND REFERENCE FRAME, in particular at the point of the trailing paddle boarder, PARTICLE B. To the eyes of PARTICLE O, both of these paddle boarders, which we will refer to as PARTICLES from here, are moving across your frame of reference at some RELATIVE VELOCITY.
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The POSITION of these particles RELATIVE to your FIXED LOCATION can be illustrated as two POSITION VECTORS, π" and π# , illustrated as:
Both of these POSITION VECTORS originate from an ORIGIN that is FIXED on earth, and because of this, they are referred to as the ABSOLUTE POSITIONS. From here, we have established the information we need to begin our analysis of the MOTION of PARTICLE A from a FIXED POINT OF REFERENCE at PARTICLE O. However, we know that there is a second set of observations occurring, which are those of PARTICLE B looking at PARTICLE Aβ¦the trailing paddle boarder observing the MOTION of the paddle boarder in front. The SECONDARY REFERENCE FRAME is fixed on PARTICLE B, but that PARTICLE is MOVING, so in the same way, the REFERENCE FRAME will also move at the same rate.
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This REFERENCE FRAME is PARALLEL to our FIXED REFERENCE FRAME on earth and since the MOTION of this frame is TRANSLATING, it is referred as the TRANSLATING FRAME:
In it important to NOTE that TRANSLATION here is used in contrast to ROTATING, or ROTATIONAL MOTION. Now from this TRANSLATING FRAME, the POSITION of PARTICLE A relative to PARTICLE B can be illustrated as the POSITION VECTOR, π"/# , which can be stated as the RELATIVE POSITION of PARTICLE A relative to the observation taking palce at PARTICLE B. From this point, we can use our knowledge of VECTOR ADDITION and the PARALLELOGRAM LAW to establish the RELATIONSHIPS between all the PARTICLES in the two REFERENCE FRAMES.
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For more work in the TOPIC of VECTORS, reference the MATHEMATICS section, more specifically the VECTOR PROPERTIES CONCEPT OVERVIEW. Other relevant information regarding VECTORS can also be referenced under the SUBJECT of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To set the POSITION of of PARTICLE A relative to PARTICLE B, we can deploy our knowledge of VECTOR SUBTRACTION to get: π"/# = π" β π# This can be rewritten as it is presented to us in the NCEES Reference Handbook as: π" = π# + π"/# With this relationship established, we can take the first TIME derivative to establish the expression for the RELATIVE VELOCITY, which is written as: π£" = π£# + π£"/# Where: π£"/# = π Γ π"/# This DOES NOT mean that we are taking the ANGULAR VELOCITY, π, and multiplying it by the RELATIVE POSITION VECTOR, π"/# . Made with by Prepineer | Prepineer.com
Rather, this expression is telling us that we are taking the CROSS PRODUCT of the ANGULAR VELOCITY, π, and the RELATIVE POSITION VECTOR, π"/# . For more work in the TOPIC of CROSS PRODUCT, reference the MATHEMATICS section, more specifically the CROSS PRODUCT CONCEPT OVERVIEW. Taking the first TIME derivative of the RELATIVE VELOCITY, or the second TIME derivative of RELATIVE POSITION, we find that: π" = π# + π"/# It is IMPORTANT to NOTE that all of these expressions are given in VECTOR FORM and only hold true when we are dealing with a FIXED and TRANSLATING FRAME OF REFERENCE of PARTICLES. They do not apply when discussing PARTICLE MOTION presiding in a ROTATING FRAME OF REFERENCE, which will be discussed when working with RIGID BODY KINEMATICS.
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RELATIVE MOTION | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
PROBLEM: Object A is traveling along a circular path with a velocity of 5 m/s that is increasing at a rate of 12 m/s 2 , as illustrated. If object B is traveling along a straight path with a speed that is decreasing at a rate of 5 m/s 2 , the relative velocity of object A is best represented as: m/s
A. 3π B. 8π + 5π C. 5π + 8π D. 50π + 12π
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SOLUTION:
The TOPIC of RELATIVE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 and 73 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RELATIVE MOTION is the calculation of the MOTION of an OBJECT WITH REGARD TO SOME OTHER MOVING OBJECT, or stationary object for that matter. The motion is not calculated with reference to the earth, but the velocity of the object in reference to the other moving object as if it were in a static state. RELATIVE MOTION is a concept, and its calculation occurs in the context of RELATIVE VELOCITY, RELATIVE SPEED, or RELATIVE ACCELERATION. In our problem, we are given a pair of objects that are traveling along two independent tracks. Each object is defined using a combination of GEOMETRY, VELOCITY, and ACCELERATION:
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We have TWO MOVING OBJECTS, which means we will be dealing with a TRANSLATING FRAME moving within a FIXED FRAME, which will be the PLANE of the page. Since it is the RELATIVE VELOCITY of OBJECT A that we are concerned with in this problem, the TRANSLATING FRAME will sit on top of OBJECT B, such that:
Based on the problem statement, and the illustration given, we have the following data defined: Object A (Purple): π£" = 5 m/s π" = 12 m/s 2 Object B (Orange): π£# = 8 m/s π# = β5 m/s 2
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These measurements are all RELATIVE to our FIXED FRAME, hence the usage of the VARIABLES π£" , π£# β¦etc. Again, we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. In VARIABLE terms, we are looking to define π£"/# . Flipping over to PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and specifically, the SECTION titled RELATIVE MOTION, we can pull the formulas that we will be using to derive this measurement. The RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B can be found using: π£" = π£# + π£"/# It is IMPORTANT to NOTE before moving forward in our computations, that this expression is given in VECTOR FORM and only holds true when we are dealing with a FIXED and TRANSLATING FRAME OF REFERENCE of PARTICLES. We are working with TRANSLATING FRAMES OF REFERENCE, so we are good there, but we donβt have our measurements in VECTOR FORM at this point, so letβs convert them in to CARTESIAN FORM. For OBJECT A, the VELOCITY is headed in the POSITIVE Y-DIRECTION, or rather: π£" = 5π m/s
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We are told that the ACCELERATION at this point is INCREASING, or that it will be in the same DIRECTION as the VELOCITY. However, for an OBJECT traveling along the path of a CURVE, the ACCLERATION will be broken up in to TWO COMPONENTS, the NORMAL and TANGENTIAL ACCELERATION. At this point in the OBJECTS travel, or any point for that matter along a curved path, we can define a set of AXES coming directly from the OBJECT itself. These AXES are referred to as the TANGENTIAL AXIS (T-AXIS), which is TANGENT to the ARC, and the NORMAL AXIS (N-AXIS), which is pointing towards the CENTER OF CURVATURE and is NORMAL to the ARC. To illustrate this, generally we will have:
At any given time, we can set up a pair of AXIS, the TANGENTIAL and NORMAL AXIS, at the POSITION of this PARTICLE.
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It will always hold true that the TANGENTIAL AXIS will run TANGENT to the CURVE at the POSITION of the OBJECT and always points in the DIRECTION of the MOTION. The NORMAL AXIS will run NORMAL to the CURVE at the POSITION of the OBJECT and always points inward towards the CENTER OF CURVATURE.
In formulaic terms, the TANGENTIAL ACCLERATION is expressed as:
π8 = π£ =
ππ£ ππ‘
The NORMAL ACCLERATION is expressed as: π£8 2 π; = π Where: π£8 = MAGNITUDE of VELOCITY at this POSITION π = INSTANTANEOUS RADIUS OF CURVATURE
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With the TANGENTIAL ACCELERATION and the NORMAL ACCELERATION components PERPENDICULAR to one another, we can express the MAGNITUDE of the OBJECTβS ACCELERATION at this POSITION as:
π=
π8 2 + π; 2
So with all that being stated, expressing the ACCELERATION in terms of the TANGENTIAL and NORMAL COMPONENTS, we get: π8 = 12 m/s 2 And: 5 m/s π; = .5
2
Or: π; = 50 m/s 2 We need to also express these values in CARTESIAN VECTOR FORM, which gives us: π8 = 12π m/s 2 And: π; = 10π m/s 2
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The ACCELERATION at any point on a curve will be expressed in terms of its NORMAL and TANGENTIAL COMPONENTS, such as for our case: π = 50π + 12π m/s 2 OBJECT B renders a much cleaner and quicker route to defining its VELOCITY and ACCELERATION in CARTESIAN VECTOR form since we arenβt dealing with any CURVES, just STRAIGHT LINE MOTION. However, the DIRECTIONS of each COMPONENT will be important to NOTE and if applied incorrectly, will be catastrophic to the conclusion of your computations. For OBJECT B, we know that it has some VELOCITY going to the LEFT, therefore: π£# = β8π m/s We also know that OBJECT B is DECCELERATING at a rate of 5 m/s 2 , or: π# = 5π m/s 2 NOTE that the HORIZONTAL COMPONENT for ACCELERATION is denoted as a POSITIVE VALUE. This is because of the fact that the OBJECT is DECCELERATING, or in other words, that the RATE OF SPEED is decreasing in an OPPOSITE DIRECTION of the MOTION of the OBJECT. Since the VELOCITY of our OBJECT in going to the LEFT, our ACCELERATION will be going in the opposite direction, to the RIGHTβ¦or the POSITIVE X (j) DIRECTION,
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Alright, so with all that being stated, we have the following, all in CARTESIAN VECTOR FORM: OBJECT A: π£" = 5π m/s π" = 50π + 12π m/s 2 OBJECT B: π£# = β8π m/s π# = 5π m/s 2 Recall that we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. The GENERAL FORMULA presented to us in the NCEES Reference Handbook for RELATIVE VELOCITY is: π£" = π£# + π£"/# Plugging in the VECTORS we have set up, we get: 5π m/s = β8π m/s + π£"/#
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Rearranging to solve for the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B, we have: π£"/# = 8π m/s + 5π m/s Or in more friendly terms: π£"/# = 8π + 5π m/s The correct answer choice is B. ππ + ππ π¦/π¬
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The TOPIC of RELATIVE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 and 73 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: When we are analyzing OBJECTS, or PARTICLES, that are in MOTION, we inherently start with some reference frame. Whether we set this frame consciously or subconsciously, all the measurements are drawn based on that particular point of reference. Our daily lives are lived within the first person REFERENCE FRAMEβ¦this is where we by default define the characteristics of MOTION such as VELOCITY and ACCELERATION, more accurately described as RELATIVE VELOCITY and RELATIVE ACCELERATION, as we will define. These characteristic measurements of motion can seem dramatically different from another individualβs perspective because their REFERENCE FRAME isnβt set at the same point as yours.
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To put an illustration around this, say that you are standing on the beaches of Kauai gazing in to the water at a couple on paddle boarders. You have your eyes set on the person in front as they are paddling across your vision at some rate of speed. The person on the second paddle board, trailing the individual in front, is also watching the leading individual as they paddle themselves to keep up. Simple observation tells us that the motion of the individual in front paddling will look different to us standing on that stagnant shore than it does to the individual trailing behind paddling at some rate of speed as well. To the individual trailing the person in front, the motion of that leading paddle boarder looks much slower than to the eyes of you who are standing still on the shore. These observations are the result of RELATIVE MOTION. RELATIVE MOTION is the calculation of the MOTION of an OBJECT WITH REGARD TO SOME OTHER MOVING OBJECT, or stationary object for that matter. The motion is not calculated with reference to the earth, but the velocity of the object in reference to the other moving object as if it were in a static state. RELATIVE MOTION is a concept, and its calculation occurs in the context of RELATIVE VELOCITY, RELATIVE SPEED, or RELATIVE ACCELERATION.
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Letβs take this illustration and set up visual context around viewing and analyzing RELATIVE MOTION problems. Since you are standing on the beach, not moving, we can establish your point of reference as a FIXED REFERENCE FRAME, such that:
For simplification purposes, you as the initial point of reference will be considered the ORIGIN, or PARTCILE O, from this point moving forward.
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With this ORIGIN established, we will define two paddle boarders off in the distance, within a SECOND REFERENCE FRAME. The paddle boarder in the back will be PARTICLE B and the paddle boarder in the front will be PARTICLE A, as such:
To properly analyze this scenario, we must set up this SECOND REFERENCE FRAME, in particular at the point of the trailing paddle boarder, PARTICLE B. To the eyes of PARTICLE O, both of these paddle boarders, which we will refer to as PARTICLES from here, are moving across your frame of reference at some RELATIVE VELOCITY.
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The POSITION of these particles RELATIVE to your FIXED LOCATION can be illustrated as two POSITION VECTORS, π" and π# , illustrated as:
Both of these POSITION VECTORS originate from an ORIGIN that is FIXED on earth, and because of this, they are referred to as the ABSOLUTE POSITIONS. From here, we have established the information we need to begin our analysis of the MOTION of PARTICLE A from a FIXED POINT OF REFERENCE at PARTICLE O. However, we know that there is a second set of observations occurring, which are those of PARTICLE B looking at PARTICLE Aβ¦the trailing paddle boarder observing the MOTION of the paddle boarder in front. The SECONDARY REFERENCE FRAME is fixed on PARTICLE B, but that PARTICLE is MOVING, so in the same way, the REFERENCE FRAME will also move at the same rate.
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This REFERENCE FRAME is PARALLEL to our FIXED REFERENCE FRAME on earth and since the MOTION of this frame is TRANSLATING, it is referred as the TRANSLATING FRAME:
In it important to NOTE that TRANSLATION here is used in contrast to ROTATING, or ROTATIONAL MOTION. Now from this TRANSLATING FRAME, the POSITION of PARTICLE A relative to PARTICLE B can be illustrated as the POSITION VECTOR, π"/# , which can be stated as the RELATIVE POSITION of PARTICLE A relative to the observation taking palce at PARTICLE B. From this point, we can use our knowledge of VECTOR ADDITION and the PARALLELOGRAM LAW to establish the RELATIONSHIPS between all the PARTICLES in the two REFERENCE FRAMES.
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For more work in the TOPIC of VECTORS, reference the MATHEMATICS section, more specifically the VECTOR PROPERTIES CONCEPT OVERVIEW. Other relevant information regarding VECTORS can also be referenced under the SUBJECT of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To set the POSITION of of PARTICLE A relative to PARTICLE B, we can deploy our knowledge of VECTOR SUBTRACTION to get: π"/# = π" β π# This can be rewritten as it is presented to us in the NCEES Reference Handbook as: π" = π# + π"/# With this relationship established, we can take the first TIME derivative to establish the expression for the RELATIVE VELOCITY, which is written as: π£" = π£# + π£"/# Where: π£"/# = π Γ π"/# This DOES NOT mean that we are taking the ANGULAR VELOCITY, π, and multiplying it by the RELATIVE POSITION VECTOR, π"/# . Made with by Prepineer | Prepineer.com
Rather, this expression is telling us that we are taking the CROSS PRODUCT of the ANGULAR VELOCITY, π, and the RELATIVE POSITION VECTOR, π"/# . For more work in the TOPIC of CROSS PRODUCT, reference the MATHEMATICS section, more specifically the CROSS PRODUCT CONCEPT OVERVIEW. Taking the first TIME derivative of the RELATIVE VELOCITY, or the second TIME derivative of RELATIVE POSITION, we find that: π" = π# + π"/# It is IMPORTANT to NOTE that all of these expressions are given in VECTOR FORM and only hold true when we are dealing with a FIXED and TRANSLATING FRAME OF REFERENCE of PARTICLES. They do not apply when discussing PARTICLE MOTION presiding in a ROTATING FRAME OF REFERENCE, which will be discussed when working with RIGID BODY KINEMATICS.
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RELATIVE MOTION | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
PROBLEM: Object A is traveling along a circular path with a velocity of 5 m/s that is increasing at a rate of 12 m/s 2 , as illustrated. If object B is traveling along a straight path with a speed that is decreasing at a rate of 5 m/s 2 , the relative velocity of object A is best represented as: m/s
A. 3π B. 8π + 5π C. 5π + 8π D. 50π + 12π
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SOLUTION:
The TOPIC of RELATIVE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 and 73 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RELATIVE MOTION is the calculation of the MOTION of an OBJECT WITH REGARD TO SOME OTHER MOVING OBJECT, or stationary object for that matter. The motion is not calculated with reference to the earth, but the velocity of the object in reference to the other moving object as if it were in a static state. RELATIVE MOTION is a concept, and its calculation occurs in the context of RELATIVE VELOCITY, RELATIVE SPEED, or RELATIVE ACCELERATION. In our problem, we are given a pair of objects that are traveling along two independent tracks. Each object is defined using a combination of GEOMETRY, VELOCITY, and ACCELERATION:
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We have TWO MOVING OBJECTS, which means we will be dealing with a TRANSLATING FRAME moving within a FIXED FRAME, which will be the PLANE of the page. Since it is the RELATIVE VELOCITY of OBJECT A that we are concerned with in this problem, the TRANSLATING FRAME will sit on top of OBJECT B, such that:
Based on the problem statement, and the illustration given, we have the following data defined: Object A (Purple): π£" = 5 m/s π" = 12 m/s 2 Object B (Orange): π£# = 8 m/s π# = β5 m/s 2
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These measurements are all RELATIVE to our FIXED FRAME, hence the usage of the VARIABLES π£" , π£# β¦etc. Again, we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. In VARIABLE terms, we are looking to define π£"/# . Flipping over to PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and specifically, the SECTION titled RELATIVE MOTION, we can pull the formulas that we will be using to derive this measurement. The RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B can be found using: π£" = π£# + π£"/# It is IMPORTANT to NOTE before moving forward in our computations, that this expression is given in VECTOR FORM and only holds true when we are dealing with a FIXED and TRANSLATING FRAME OF REFERENCE of PARTICLES. We are working with TRANSLATING FRAMES OF REFERENCE, so we are good there, but we donβt have our measurements in VECTOR FORM at this point, so letβs convert them in to CARTESIAN FORM. For OBJECT A, the VELOCITY is headed in the POSITIVE Y-DIRECTION, or rather: π£" = 5π m/s
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We are told that the ACCELERATION at this point is INCREASING, or that it will be in the same DIRECTION as the VELOCITY. However, for an OBJECT traveling along the path of a CURVE, the ACCLERATION will be broken up in to TWO COMPONENTS, the NORMAL and TANGENTIAL ACCELERATION. At this point in the OBJECTS travel, or any point for that matter along a curved path, we can define a set of AXES coming directly from the OBJECT itself. These AXES are referred to as the TANGENTIAL AXIS (T-AXIS), which is TANGENT to the ARC, and the NORMAL AXIS (N-AXIS), which is pointing towards the CENTER OF CURVATURE and is NORMAL to the ARC. To illustrate this, generally we will have:
At any given time, we can set up a pair of AXIS, the TANGENTIAL and NORMAL AXIS, at the POSITION of this PARTICLE.
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It will always hold true that the TANGENTIAL AXIS will run TANGENT to the CURVE at the POSITION of the OBJECT and always points in the DIRECTION of the MOTION. The NORMAL AXIS will run NORMAL to the CURVE at the POSITION of the OBJECT and always points inward towards the CENTER OF CURVATURE.
In formulaic terms, the TANGENTIAL ACCLERATION is expressed as:
π8 = π£ =
ππ£ ππ‘
The NORMAL ACCLERATION is expressed as: π£8 2 π; = π Where: π£8 = MAGNITUDE of VELOCITY at this POSITION π = INSTANTANEOUS RADIUS OF CURVATURE
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With the TANGENTIAL ACCELERATION and the NORMAL ACCELERATION components PERPENDICULAR to one another, we can express the MAGNITUDE of the OBJECTβS ACCELERATION at this POSITION as:
π=
π8 2 + π; 2
So with all that being stated, expressing the ACCELERATION in terms of the TANGENTIAL and NORMAL COMPONENTS, we get: π8 = 12 m/s 2 And: 5 m/s π; = .5
2
Or: π; = 50 m/s 2 We need to also express these values in CARTESIAN VECTOR FORM, which gives us: π8 = 12π m/s 2 And: π; = 10π m/s 2
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The ACCELERATION at any point on a curve will be expressed in terms of its NORMAL and TANGENTIAL COMPONENTS, such as for our case: π = 50π + 12π m/s 2 OBJECT B renders a much cleaner and quicker route to defining its VELOCITY and ACCELERATION in CARTESIAN VECTOR form since we arenβt dealing with any CURVES, just STRAIGHT LINE MOTION. However, the DIRECTIONS of each COMPONENT will be important to NOTE and if applied incorrectly, will be catastrophic to the conclusion of your computations. For OBJECT B, we know that it has some VELOCITY going to the LEFT, therefore: π£# = β8π m/s We also know that OBJECT B is DECCELERATING at a rate of 5 m/s 2 , or: π# = 5π m/s 2 NOTE that the HORIZONTAL COMPONENT for ACCELERATION is denoted as a POSITIVE VALUE. This is because of the fact that the OBJECT is DECCELERATING, or in other words, that the RATE OF SPEED is decreasing in an OPPOSITE DIRECTION of the MOTION of the OBJECT. Since the VELOCITY of our OBJECT in going to the LEFT, our ACCELERATION will be going in the opposite direction, to the RIGHTβ¦or the POSITIVE X (j) DIRECTION,
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Alright, so with all that being stated, we have the following, all in CARTESIAN VECTOR FORM: OBJECT A: π£" = 5π m/s π" = 50π + 12π m/s 2 OBJECT B: π£# = β8π m/s π# = 5π m/s 2 Recall that we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. The GENERAL FORMULA presented to us in the NCEES Reference Handbook for RELATIVE VELOCITY is: π£" = π£# + π£"/# Plugging in the VECTORS we have set up, we get: 5π m/s = β8π m/s + π£"/#
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Rearranging to solve for the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B, we have: π£"/# = 8π m/s + 5π m/s Or in more friendly terms: π£"/# = 8π + 5π m/s The correct answer choice is B. ππ + ππ π¦/π¬
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