Question

A particle undergoes a constant acceleration of 4.00 m/s2. After a certain amount of time, its velocity is 12.2 m/s. (Where applicable, indicate the direction with the sign of your answer.) (a) If its initial velocity is 6.1 m/s, what is its displacement during this time? m (b) What distance does it travel during this time? m (c) If its initial velocity is −6.1 m/s, what is its displacement during this time? m (d) What is the total distance the particle travels during the interval in part (c)? m

Answer 1

(a)

initial velocity vox = 6.1 m/s

acceleration ax = 4 m/s^2

final velocity vx = 12.2 m/s

v = vox + ax*t

12.2 = 6.1 + 4*t

t = 1.525 s

x = vox*t + (1/2)*ax*t^2

x = (6.1*1.525) + (1/2)*4*1.525^2

x = 13.95 m

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(b)

the velocity and acceleration are positive , so the particel moves

along +x axis during the time

the distance = displacement = 13.95 m

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(c)

initial velocity vox = 6.1 m/s

acceleration ax = 4 m/s^2

final velocity vx = 12.2 m/s

v = vox + ax*t

12.2 = -6.1 + 4*t

t = 4.575 s

x = vox*t + (1/2)*ax*t^2

x = -(6.1*4.575) + (1/2)*4*4.575^2

x = 13.95 m

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(d)

the velocity and acceleration are opposite sign , so the particle moves

along -x axis with decreasing speed , comes to rest at instant of time

and then moves along +x axis and then acquires vx

at some time t velocity = 0

vf = vox + a*t

0 = -6.1 + (4*t)

t = 1.525 s the speed is zero

displacement travelled = -6.1*1.525 + ((1/2)*4*1.515^2) = -4.65 m

distance = 4.65 m

from t = 1.525 to t = 4.575 s

displacement travelled = (0*(4.575-1.525)) + ( (1/2)*4*(4.57-1.525)^2) = 18.5 m

total distance = 4.65+18.5 = 23.15 m