Question

A car starts from rest, attains a velocity of $36km{h}^{-1}$ with an acceleration of $0.2m{s}^{-2}$, travels 9km with this uniform velocity and then comes to halt with a uniform deceleration of $0.1m{s}^{-2}$. The total time of travel of the car is

• A. 1050 s
• B. 1000 s
• C. 950 s
• D. 900 s

Answer 1

Answer: (A)

1050 s

Given

$v=36km{h}^{-1}=\frac{36000}{3600}=10m{s}^{-1}$

$s=9km=9000m$

This motion is in three phases.

[Case: 1]
with acceleration, $a=0.2m{s}^{-2}$ and intial velocity , $u=0$

Thus using equation $v=u+at$

$10=0+0.2\times t$

So, $t=50s$

[Case: 2]
uniform motion with velocity ,$v=10$
and distance, $s=9000$

Thus time , $t=\frac{s}{v}=\frac{9000}{10}=900s$

[Case: 3]
With acceleration, $a=-0.1m{s}^{-2}$
$u=10m{s}^{-1}$
and $v=0$

Thus using equation $v=u+at$

$0=10-0.1\times t$

So, $t=\frac{10}{0.1}=100s$

Therefore

Total time = $50s+900s+100s=1050s$

So, option $A$ is correct.