Question

a car moving with a velocity of 30 m/s is stopped by the application of breaks which impart a retardation ( negative acceleration) of 6 m/s² to the car.(1)how long does it take to come to rest(2)how far does the car travel during the time brakes are applied

Answer 1

u=30m/s
v=0
a= -6m/s²
a=v-u/t
t=v-u/a
=0-30/-6
5sec

s=ut+½at²
=30×5 - ½×6×5×5
=150-75
=75m

Answer 2

Concept:

Numerous dimensional motion issues that take into account the motion of an object with constant acceleration can be solved using the kinematic equations. The first and third kinematic equation is expressed as, v = u + at and v² = u²+ 2as, respectively

Given:

initial velocity, u  = 30 m/s

Retarding acceleration, a = -6m/s²

Final velocity, v = 0m/s

Find:

We need to determine the time period, t that the car takes to come to rest and the distance, s travelled by car during the time the brakes are applied.

Solution:

Numerous dimensional motion issues that take into account the motion of an object with constant acceleration can be solved using the kinematic equations. There are three kinematical equations for accelerated motion in one dimension. Out of those three, the first and  third kinematic equation is expressed as,

v = u + at and v² = u²+ 2as where v = final velocity, u = initial velocity, a = acceleration and s = displacement

To calculate the time period, the first kinematical equation is applied

v = u + at

0 = 30 + (-6)t

6t = 30

Therefore, t = 30/6 = 5s

The car takes about 5s to come to rest.

To calculate the distance, apply third kinematical equation

v² = u²+ 2as

0² = 30² + 2(-6)s

12s = 900

s = 900/12

s  = 75m

The distance travelled by car during the time brakes are applied is s=75m

Thus, (1) The car takes about 5s to come to rest.

(2) The car travelled about 75m during the time brakes are applied.

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