Part A What are the ball\'s average velocity during the first 3.0 s ? Part B Wha
ID: 2002486 • Letter: P
Question
Part A
What are the ball's average velocity during the first 3.0 s ?
Part B
What are the ball's average speed during the first 3.0 s ?
Express your answer using two significant figures.
Part C
Suppose that the ball moved in such a way that the graph segment after 2.0 swas -3.0 m/s instead of +3.0 m/s. Find the ball's and average velocity during the first 3.0 s in this case.
Express your answer using two significant figures.
Part D
Suppose that the ball moved in such a way that the graph segment after 2.0 swas -3.0 m/s instead of +3.0 m/s. Find the ball's average speed during the first 3.0 s in this case.
Express your answer using two significant figures.
Explanation / Answer
Solution:
Part (A)
The ball has velocity of v1 = +2 m/s from 0 s to 2 s; t1 = (2-0)s = 2 s
The displacement of the ball during 0 to 2 seconds is given by,
x1 = v1*t1
x1 = (+2m/s)*(2s)
x1 = +4 m
The ball has velocity of v2 = +3 m/s from 2 s to 3 s; t2 = (3-2)s = 1 s
The displacement of the ball during 2 to 3 seconds is given by,
x2 = v2*t2
x2 = (+3m/s)*(1s)
x2 = +3 m
The net displacement during the t = t1 + t3 = 2s +1s = 3s is,
x = x1 + x2
x = (+4 m) + (+3m)
x = +7 m
The average velocity during the t = 3s is,
vavg = displacement/time
vavg = x/t
vavg = (+7m)/(3s)
vavg = +2.3 m/s
or the ball’s average velocity during first 3.0 s is 2.3m/s
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Part (b)
Now total actual distance travelled by the ball during t = 3.0 s is,
l = | x1|+| x2|
l = |+4m|+|+3m| taking only modulus (positive values) for displacement,
l = 4m + 3m
l = 7 m
Its average speed is given by,
Savg = distance/time
Savg = (7m)/(3s)
Savg = 2.3 m/s
Thus the ball’s average speed during first 3.0 s is 2.3 m/s
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Part (C)
The ball has velocity of v1 = +2 m/s from 0 s to 2 s; t1 = (2-0)s = 2 s
The displacement of the ball during 0 to 2 seconds is given by,
x1 = v1*t1
x1 = (+2m/s)*(2s)
x1 = +4 m
The ball has velocity of v2 = -3 m/s from 2 s to 3 s; t2 = (3-2)s = 1 s
The displacement of the ball during 2 to 3 seconds is given by,
x2 = v2*t2
x2 = (-3m/s)*(1s)
x2 = -3 m
The net displacement during the t = t1 + t3 = 2s +1s = 3s is,
x = x1 + x2
x = (+4 m) + (-3m)
x = +1 m
The average velocity during the t = 3s is,
vavg = displacement/time
vavg = x/t
vavg = (+1m)/(3s)
vavg = +2.3 m/s
or the ball’s average velocity during first 3.0 s is 0.33m/s
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Part (D)
Now total actual distance travelled by the ball during t = 3.0 s is,
l = | x1|+| x2|
l = |+4m|+|-3m|
l = 4m + 3m
l = 7 m
Its average speed is given by,
Savg = distance/time
Savg = (7m)/(3s)
Savg = 2.3 m/s
Thus the ball’s average speed during first 3.0 s is 2.3 m/s