Chapter 24: Gauss' Law
1
Lecture 3 y Position and displacement y Average velocity and average speed y Instantaneous velocity and speed y Acceleration y Motion with constant acceleration
09/05/2006
2
Student response system transmitters assignment: Baker, Maria #1
Gazic, Bozana #2
Humenuik, Joseph #3
Itomlenskis, Matthew #4
Kitsis, Viktor #5
McKinney, Antoinne #6
Pizzo, Jacob #7
Uti, Anthony #8
Guest #9
Guest # 10
Guest # 11
3
Homework Assignment # 2 (due Tuesday, September 12th) Chapter 2: Problems: 1, 8, 12, 15, 22, 25, 39, 42, 47, 52
Reading Assignment Chapter 2 - Page 16 - 17 – Sample problems 2-1 - Page 18 – 19 – Sample problems 2-2 & 2-3 - Page 23 – Sample problem 2-5
Quiz #1 - (due Thursday, September 7th) Go to: http://edugen.wiley.com/edugen/class/cls24530/ and register
4
Review y Vector decomposition
r a x = a cos θ r a y = a sin θ r a = a 2x + a 2y
tan θ =
ay ax
r r r ˆ ˆ ˆ a = a x i + a y j = a ⋅ cosθ i + a ⋅ sinθ ˆj
ˆj
ˆi
ˆi = ˆj = 1
y Convention - the angle θ is always measured counter-clockwise from the positive x direction.
5
Review y Vector addition - geometrical (head to tail)
r b
r a
r b
r s
r a
- using components
y
a y ˆj θB
b y ˆj
y
y r a θA
y
sy = a y + by
(
)
r s = (a x + b x )ˆi + a y + b y ˆj
b x ˆj
a x ˆi
x
x r b
r r r s = a + b = (a x + b x ) ˆi + (a y + b y ) ˆj
sx = a x + bx
with
x
x
r r a x = a cos θ A ; a y = a sin θ A r r b x = b cos θ B ; b y = b sin θ B
6
Motion along a straight line (one – dimensional) y Definition: motion is defined as the change of an object’s position with time y In this chapter we will impose a series of restrictions: - motion is constrained in one dimension, i.e. along a straight line (typically along the x or y axis). - the motion can be either in the positive or in the negative direction of the axis used. - for now, we will neglect the forces (pushes or pulls) that determine an object to move). - the moving object is either a particle or an object that moves like a particle (all its points move in the same direction at the same rate (speed).
7
Position of an object y For a one-dimensional motion the position of the object is specified by a single coordinate x. y In order to describe the position of an object: - take a snapshot of the object at different times and record its position. - plot this position as a function of time x = x(t) - you might need to fit the plot to obtain the missing points.
8
Examples of time dependence of position
x (m)
x (m)
x (m)
t (s)
t (s)
t (s)
9
Displacement y A change from one position x1 to another position x2:
Δx = x 2 − x1 Note: displacement is a vector even if in this chapter we will not specify it all the time. Consequently, make sure that the sign (i.e. direction) is not ignored.
Δx = Δx ′
but
Δx ≠ Δx ′
x1′
x2
x1
x ′2
10
There are four pairs of initial and final positions. Which pair(s) gives a negative displacement: (a) –3m, 5m (b) –3m, -7m (c) 3m, -3.4m (d) –3m, 0m (A) (a) and (b) (B) (b) and (d) (C) (c) and (d) (D)
only (b)
(E)
(b) and (c)
11
For a certain interval of time, is the magnitude of the displacement always equal to the distance traveled? (A)
Yes
(B)
No
12
Average velocity y One characteristic of the motion is the rate of change of the object (particle) position. The initial and final position of the two objects are the same.
y Average velocity vavg is the ratio of
the displacement and the time interval over which it has occurred.
v avg
Δx x 2 − x1 = = Δt t 2 − t1
- Dimension is: [vavg]= L/T - Unit: m/s
13
Graphic interpretation of the velocity y The velocity is the slope of the straight line that connects the two particular points over which the displacement is calculated.
x x2
Slope of this line
x1 t1
t2
t
14
According to the following graph, when the two bodies (1) and (2) have the same velocity: x (m) (A) t = 0 s (1)
(B)
t=5s
(C)
t = 10 s
(D)
Never
(E)
None of the above
(2)
t (s) 0
5
10
15
Average speed y Average speed is the ratio of the total distance traveled (Ex. number of meters moved over the time interval
s avg
Total Distance = Δt
y Because the time interval is always positive the average speed is always positive
Example:
v avg
Δx x 2 − x1 = = Δt t 2 − t1
s avg
Total Distance = Δt
16
A car goes form city A to city B and then returns to city C (see figure). If it takes 10 minutes to drive from A to B, and 14 minutes from B to C, and the car was stationed in city B for half an hour, what are the average velocity and speed? C
A
B
Δt = t AB + t B + t BC = 600 s + 1800 s + 240 s = 2640 s
Total distance = AB + BC = 3 km + 6 km = 9 km = 9000 m Δx = x C − x A = −2 km − 1km = -3 km = - 3000 m v avg =
Δx −3000 m m = ≈ −1.14 Δt 2640 s s
s avg =
Total distance 9000 m m = ≈ 3 .4 Δt 2640 s s
Note: - typically the value of the average velocity and speed are different
17
Instantaneous velocity y To obtain the velocity at any instance, the time interval Δt over which the average velocity is calculated is reduced
Δx dx = v = lim Δt → 0 Δt dt - v is the derivative of x with respect to t.
e = ty n i i si l loc th ve f o ge e a o p v er l S a
- v is the slope of the tangent to the position versus time curve at the point of interest.
y Speed = the magnitude of instantaneous velocity. It does not contain any indication about the direction of motion.
x(t)
18
The time dependence of the position of an object is shown in the figure. At which point is the object at rest (zero velocity)? (A)
D and F
(B)
A and C
x (m)
B A C
(C)
B and E
(D)
B, D, E and F
F
t (s)
D
(E)
A, C, E and F E
19
Acceleration y Acceleration – characterizes the rate of change in the velocity of an object (particle)
a avg
Δv v 2 − v1 = = Δt t 2 − t1
v v2
- Dimension is: [aavg]=L/T2 - Unit: m/s2
v1 - aavg – is the slope of this line
t1
t2
t
20
Instantaneous Acceleration y Instantaneous acceleration – derivative of velocity with respect to time
Δv dv = a = lim Δt → 0 Δt dt
dv d ⎛ dx ⎞ d 2 x a = = ⎜ ⎟= 2 dt dt ⎝ dt ⎠ dt
21
The acceleration’s sign y A negative sign for the acceleration does not necessarily means that the speed of an object is decreasing. Ex.: - an object starts from rest and increases its speed to (– 10 m/s) in 5s. v x
a avg
Δv −10 − 0 m m = = = −2 2 2 Δt 5 s s
- the acceleration is negative even if the objects accelerates, because the motion is in the negative direction of the x axis Note: - if the sign of the velocity and acceleration of an object are the same, the speed of the object increases. If the signs are opposite the speed decreases.
x (m)
22 Problem: The position of a car versus time is described by the following graph. a) Find the displacement and total distance traveled in 60 s. b) Plot the time dependence of the velocity c) Calculate the average velocity and average speed d) Calculate the average acceleration in the time interval 0 s to 60 s. e) Plot the time dependence of the acceleration (comment on the result)
45 40 35 30 25 20 15 10 5 0 0
5
10 15 20 25 30 35 40 45 50 55 60 t (s)
23
a) Δx = 20 - 0 = 20 = 60 m 0 −5s 5 − 15 s 15 − 25 s 25 − 40 s 40 − 45 s 45 − 55 s 55 − 60 s
m 0−0 =0 s 5−0 m 10 − 0 v= =1 s 15 − 5 m 10 − 10 v= =0 s 25 − 15 m 40 − 10 v= =2 s 40 − 25 m 40 − 40 v= =0 s 45 − 40 m 20 − 40 v= = −2 s 55 − 45 m 20 − 20 v= =0 s 60 − 55
v=
25 20 15 10 5 0 0
5
10 15 20 25 30 35 40 45 50 55 60 t (s)
2.5 2 1.5 v (m/s)
b)
35 30 x (m)
Total distance = 40 - 0 + 20 − 40
45 40
1 0.5 0 -0.5 0 -1
5 10 15 20 25 30 35 40 45 50 55 60
-1.5 -2 -2.5 t (s)
c)
Δx 20m m = = 0.33 Δt 60s s Total distance 60m m = = =1 Δt 60s s
v avg = s avg
24
m Δv 0 - 0 = =0 2 Δt 60s s
0−0 m =0 2 5−0 s 1−1 m =0 2 5 − 15 s a= 15 − 5 s ............. a - will be always zero
e) 0 − 5 s
a=
Note: - the measurement of the time dependence of the position was poor and it does not completely reflect the reality.
v (m/s)
a avg =
2.5 2 1.5 1 0.5 0 -0.5 0 -1
5 10 15 20 25 30 35 40 45 50 55 60
-1.5 -2 -2.5 t (s)
0.5 0.4 0.3 a (m/s^2)
d)
0.2 0.1 0 -0.1 0 -0.2
5 10 15 20 25 30 35 40 45 50 55 60
-0.3 -0.4 -0.5 t (s)
25
2.5 2 1.5
0
5
10 15 20 25 30 35 40 45 50 55 60
1 0.5 0 -0.5 0 -1
0.5 0.4 0.3
5 10 15 20 25 30 35 40 45 50 55 60
a (m/s^2)
35 30 25 20 15 10 5 0
v (m/s)
x (m)
45 40
-1.5 -2 -2.5
0.2 0.1 0 -0.1 0 -0.2
5 10 15 20 25 30 35 40 45 50 55 60
-0.3 -0.4 -0.5
t (s)
t (s)
t (s)
More real measurement: 2.5
0.6
2 1.5
0.4
1 0.5
0.2
0 -0.5 0 -1
5
10 15 20 25 30 35 40 45 50 55 60
5
10 15 20 25 30 35 40 45 50 55 60 t (s)
0 0
5
10 15 20 25 30 35 40 45 50 55 60
-0.2
-1.5 -2
0
a (m/s^2)
35 30 25 20 15 10 5 0
v (m/s)
a (m/s^2)
45 40
-0.4
-2.5
-0.6
t (s)
t (s)
26
(A) +3 m/s (B) +2 m/s (C) -2 m/s (D) -3 m/s (E) None of the above
x(m)
What is vx at t = 1 s?
4 3 2 1 0 -1 0 -2 -3 -4
1
2
3
4
5
t (s)
6
7
8
9
10
27
(A) +3 m/s (B) +2 m/s (C) -2 m/s (D) -3 m/s (E) None of the above
x(m)
What is vx at t = 3 s?
4 3 2 1 0 -1 0 -2 -3 -4
1
2
3
4
5
t (s)
6
7
8
9
10
2 chances What is the average and t = 8 s? (A) ~ 0.43 m/s2 (B) ~ 0.21 m/s2 (C) ~ - 0.43 m/s2 (D) ~ 0 m/s2 (E) None of the above
x(m)
acceleration between t = 1s
4 3 2 1 0 -1 0 -2 -3 -4
1
2
3
4
5
t (s)
6
7
8
9
28
10
29
Constant acceleration case
y When the acceleration is constant the average acceleration and instantaneous acceleration are equal:
a = a avg
v − v0 = t −0
v = v 0 + at Eq. (1)
Lecture 3 y Position and displacement y Average velocity and average speed y Instantaneous velocity and speed y Acceleration y Motion with constant acceleration
09/05/2006
2
Student response system transmitters assignment: Baker, Maria #1
Gazic, Bozana #2
Humenuik, Joseph #3
Itomlenskis, Matthew #4
Kitsis, Viktor #5
McKinney, Antoinne #6
Pizzo, Jacob #7
Uti, Anthony #8
Guest #9
Guest # 10
Guest # 11
3
Homework Assignment # 2 (due Tuesday, September 12th) Chapter 2: Problems: 1, 8, 12, 15, 22, 25, 39, 42, 47, 52
Reading Assignment Chapter 2 - Page 16 - 17 – Sample problems 2-1 - Page 18 – 19 – Sample problems 2-2 & 2-3 - Page 23 – Sample problem 2-5
Quiz #1 - (due Thursday, September 7th) Go to: http://edugen.wiley.com/edugen/class/cls24530/ and register
4
Review y Vector decomposition
r a x = a cos θ r a y = a sin θ r a = a 2x + a 2y
tan θ =
ay ax
r r r ˆ ˆ ˆ a = a x i + a y j = a ⋅ cosθ i + a ⋅ sinθ ˆj
ˆj
ˆi
ˆi = ˆj = 1
y Convention - the angle θ is always measured counter-clockwise from the positive x direction.
5
Review y Vector addition - geometrical (head to tail)
r b
r a
r b
r s
r a
- using components
y
a y ˆj θB
b y ˆj
y
y r a θA
y
sy = a y + by
(
)
r s = (a x + b x )ˆi + a y + b y ˆj
b x ˆj
a x ˆi
x
x r b
r r r s = a + b = (a x + b x ) ˆi + (a y + b y ) ˆj
sx = a x + bx
with
x
x
r r a x = a cos θ A ; a y = a sin θ A r r b x = b cos θ B ; b y = b sin θ B
6
Motion along a straight line (one – dimensional) y Definition: motion is defined as the change of an object’s position with time y In this chapter we will impose a series of restrictions: - motion is constrained in one dimension, i.e. along a straight line (typically along the x or y axis). - the motion can be either in the positive or in the negative direction of the axis used. - for now, we will neglect the forces (pushes or pulls) that determine an object to move). - the moving object is either a particle or an object that moves like a particle (all its points move in the same direction at the same rate (speed).
7
Position of an object y For a one-dimensional motion the position of the object is specified by a single coordinate x. y In order to describe the position of an object: - take a snapshot of the object at different times and record its position. - plot this position as a function of time x = x(t) - you might need to fit the plot to obtain the missing points.
8
Examples of time dependence of position
x (m)
x (m)
x (m)
t (s)
t (s)
t (s)
9
Displacement y A change from one position x1 to another position x2:
Δx = x 2 − x1 Note: displacement is a vector even if in this chapter we will not specify it all the time. Consequently, make sure that the sign (i.e. direction) is not ignored.
Δx = Δx ′
but
Δx ≠ Δx ′
x1′
x2
x1
x ′2
10
There are four pairs of initial and final positions. Which pair(s) gives a negative displacement: (a) –3m, 5m (b) –3m, -7m (c) 3m, -3.4m (d) –3m, 0m (A) (a) and (b) (B) (b) and (d) (C) (c) and (d) (D)
only (b)
(E)
(b) and (c)
11
For a certain interval of time, is the magnitude of the displacement always equal to the distance traveled? (A)
Yes
(B)
No
12
Average velocity y One characteristic of the motion is the rate of change of the object (particle) position. The initial and final position of the two objects are the same.
y Average velocity vavg is the ratio of
the displacement and the time interval over which it has occurred.
v avg
Δx x 2 − x1 = = Δt t 2 − t1
- Dimension is: [vavg]= L/T - Unit: m/s
13
Graphic interpretation of the velocity y The velocity is the slope of the straight line that connects the two particular points over which the displacement is calculated.
x x2
Slope of this line
x1 t1
t2
t
14
According to the following graph, when the two bodies (1) and (2) have the same velocity: x (m) (A) t = 0 s (1)
(B)
t=5s
(C)
t = 10 s
(D)
Never
(E)
None of the above
(2)
t (s) 0
5
10
15
Average speed y Average speed is the ratio of the total distance traveled (Ex. number of meters moved over the time interval
s avg
Total Distance = Δt
y Because the time interval is always positive the average speed is always positive
Example:
v avg
Δx x 2 − x1 = = Δt t 2 − t1
s avg
Total Distance = Δt
16
A car goes form city A to city B and then returns to city C (see figure). If it takes 10 minutes to drive from A to B, and 14 minutes from B to C, and the car was stationed in city B for half an hour, what are the average velocity and speed? C
A
B
Δt = t AB + t B + t BC = 600 s + 1800 s + 240 s = 2640 s
Total distance = AB + BC = 3 km + 6 km = 9 km = 9000 m Δx = x C − x A = −2 km − 1km = -3 km = - 3000 m v avg =
Δx −3000 m m = ≈ −1.14 Δt 2640 s s
s avg =
Total distance 9000 m m = ≈ 3 .4 Δt 2640 s s
Note: - typically the value of the average velocity and speed are different
17
Instantaneous velocity y To obtain the velocity at any instance, the time interval Δt over which the average velocity is calculated is reduced
Δx dx = v = lim Δt → 0 Δt dt - v is the derivative of x with respect to t.
e = ty n i i si l loc th ve f o ge e a o p v er l S a
- v is the slope of the tangent to the position versus time curve at the point of interest.
y Speed = the magnitude of instantaneous velocity. It does not contain any indication about the direction of motion.
x(t)
18
The time dependence of the position of an object is shown in the figure. At which point is the object at rest (zero velocity)? (A)
D and F
(B)
A and C
x (m)
B A C
(C)
B and E
(D)
B, D, E and F
F
t (s)
D
(E)
A, C, E and F E
19
Acceleration y Acceleration – characterizes the rate of change in the velocity of an object (particle)
a avg
Δv v 2 − v1 = = Δt t 2 − t1
v v2
- Dimension is: [aavg]=L/T2 - Unit: m/s2
v1 - aavg – is the slope of this line
t1
t2
t
20
Instantaneous Acceleration y Instantaneous acceleration – derivative of velocity with respect to time
Δv dv = a = lim Δt → 0 Δt dt
dv d ⎛ dx ⎞ d 2 x a = = ⎜ ⎟= 2 dt dt ⎝ dt ⎠ dt
21
The acceleration’s sign y A negative sign for the acceleration does not necessarily means that the speed of an object is decreasing. Ex.: - an object starts from rest and increases its speed to (– 10 m/s) in 5s. v x
a avg
Δv −10 − 0 m m = = = −2 2 2 Δt 5 s s
- the acceleration is negative even if the objects accelerates, because the motion is in the negative direction of the x axis Note: - if the sign of the velocity and acceleration of an object are the same, the speed of the object increases. If the signs are opposite the speed decreases.
x (m)
22 Problem: The position of a car versus time is described by the following graph. a) Find the displacement and total distance traveled in 60 s. b) Plot the time dependence of the velocity c) Calculate the average velocity and average speed d) Calculate the average acceleration in the time interval 0 s to 60 s. e) Plot the time dependence of the acceleration (comment on the result)
45 40 35 30 25 20 15 10 5 0 0
5
10 15 20 25 30 35 40 45 50 55 60 t (s)
23
a) Δx = 20 - 0 = 20 = 60 m 0 −5s 5 − 15 s 15 − 25 s 25 − 40 s 40 − 45 s 45 − 55 s 55 − 60 s
m 0−0 =0 s 5−0 m 10 − 0 v= =1 s 15 − 5 m 10 − 10 v= =0 s 25 − 15 m 40 − 10 v= =2 s 40 − 25 m 40 − 40 v= =0 s 45 − 40 m 20 − 40 v= = −2 s 55 − 45 m 20 − 20 v= =0 s 60 − 55
v=
25 20 15 10 5 0 0
5
10 15 20 25 30 35 40 45 50 55 60 t (s)
2.5 2 1.5 v (m/s)
b)
35 30 x (m)
Total distance = 40 - 0 + 20 − 40
45 40
1 0.5 0 -0.5 0 -1
5 10 15 20 25 30 35 40 45 50 55 60
-1.5 -2 -2.5 t (s)
c)
Δx 20m m = = 0.33 Δt 60s s Total distance 60m m = = =1 Δt 60s s
v avg = s avg
24
m Δv 0 - 0 = =0 2 Δt 60s s
0−0 m =0 2 5−0 s 1−1 m =0 2 5 − 15 s a= 15 − 5 s ............. a - will be always zero
e) 0 − 5 s
a=
Note: - the measurement of the time dependence of the position was poor and it does not completely reflect the reality.
v (m/s)
a avg =
2.5 2 1.5 1 0.5 0 -0.5 0 -1
5 10 15 20 25 30 35 40 45 50 55 60
-1.5 -2 -2.5 t (s)
0.5 0.4 0.3 a (m/s^2)
d)
0.2 0.1 0 -0.1 0 -0.2
5 10 15 20 25 30 35 40 45 50 55 60
-0.3 -0.4 -0.5 t (s)
25
2.5 2 1.5
0
5
10 15 20 25 30 35 40 45 50 55 60
1 0.5 0 -0.5 0 -1
0.5 0.4 0.3
5 10 15 20 25 30 35 40 45 50 55 60
a (m/s^2)
35 30 25 20 15 10 5 0
v (m/s)
x (m)
45 40
-1.5 -2 -2.5
0.2 0.1 0 -0.1 0 -0.2
5 10 15 20 25 30 35 40 45 50 55 60
-0.3 -0.4 -0.5
t (s)
t (s)
t (s)
More real measurement: 2.5
0.6
2 1.5
0.4
1 0.5
0.2
0 -0.5 0 -1
5
10 15 20 25 30 35 40 45 50 55 60
5
10 15 20 25 30 35 40 45 50 55 60 t (s)
0 0
5
10 15 20 25 30 35 40 45 50 55 60
-0.2
-1.5 -2
0
a (m/s^2)
35 30 25 20 15 10 5 0
v (m/s)
a (m/s^2)
45 40
-0.4
-2.5
-0.6
t (s)
t (s)
26
(A) +3 m/s (B) +2 m/s (C) -2 m/s (D) -3 m/s (E) None of the above
x(m)
What is vx at t = 1 s?
4 3 2 1 0 -1 0 -2 -3 -4
1
2
3
4
5
t (s)
6
7
8
9
10
27
(A) +3 m/s (B) +2 m/s (C) -2 m/s (D) -3 m/s (E) None of the above
x(m)
What is vx at t = 3 s?
4 3 2 1 0 -1 0 -2 -3 -4
1
2
3
4
5
t (s)
6
7
8
9
10
2 chances What is the average and t = 8 s? (A) ~ 0.43 m/s2 (B) ~ 0.21 m/s2 (C) ~ - 0.43 m/s2 (D) ~ 0 m/s2 (E) None of the above
x(m)
acceleration between t = 1s
4 3 2 1 0 -1 0 -2 -3 -4
1
2
3
4
5
t (s)
6
7
8
9
28
10
29
Constant acceleration case
y When the acceleration is constant the average acceleration and instantaneous acceleration are equal:
a = a avg
v − v0 = t −0
v = v 0 + at Eq. (1)
