Question

A car starts from rest. It accelerates uniformly at the rate of $5m{s}^{-2}$ and achieves a maximum velocity of $40m{s}^{-1}$. It then decelerates at the rate of $10m{s}^{-2}$ and stops. Find the total time of the car’s journey.

• A. 8 s
• B. 10 s
• C. 12 s
• D. 16 s

Answer 1

The correct option is C 12 s

Let’s divide the journey into two parts.
1. Acceleration from $0m{s}^{-1}$ to $40m{s}^{-1}$
2. Deceleration from $40m{s}^{-1}$ to $0m{s}^{-1}$

Let's find the time separately for each part and add them to get the total time.

We know that, $a=\frac{v-u}{t}$
⇒ $t=\frac{v-u}{a}$

For the first part of the journey,
$v=40m{s}^{-1}$
$u=0$
$a=5m{s}^{-2}$

Time taken to accelerate,
⇒ $t_1=\frac{v-u}{a}$
⇒ $t_1=\frac{40-0}{5}$
⇒ $t_1=8s$

For the second part of the journey,
$v=0$
$u=40m{s}^{-1}$
$a=-10m{s}^{-2}$ (deceleration)

Time taken to accelerate,
⇒ $t_2=\frac{0-40}{-10}$
⇒ $t_2=\frac{-40}{-10}$
⇒ $t_2=4s$

Total time $=t_1+t_2=8s+4s=12s$