Question

A vehicle can accelerate or decelerate at a maximum value of $1m{s}^{-2}$ and can attain a maximum speed of $20m{s}^{-1}$. If it starts from rest, what is the shortest time in which it can travel $1000m$, in a straight path, if it has to come to rest at the end of $1000m$?

• A. $50s$
• B. $70s$
• C. $30s$
• D. $20s$

Answer 1

The correct option is B $70s$

From the question we know that the vehicle starts from rest, attains a maximum speed of $20m/s$, and then comes to rest at the end.
This means the vehicle must have first accelerated to the maximum velocity and then decelerated to rest.

Let $t_1$ and $t_2$ be the time for acceleration (attaining maximum velocity) and deceleration (come to the rest) respectively.
Then, $t_1=\frac{\text{Maximum velocity-Initial velocity}}{\text{acceleration}}$
$=\frac{20-0}{1}=20s$
Then, distance travelled in time $t_1$,
$S_1=u{t}_1+\frac{1}{2}a{t}_1^2=0+\frac{1}{2}\times 1\times {20}^2=200m$

Similarly, $t_2=\frac{0-20}{-1}=20s$
and $S_2=u{t}_2+\frac{1}{2}a{t}_2^2=20\times 20-\frac{1}{2}\times 1\times {20}^2=200\text{m}$

$\therefore$ Distance travelled at uniform velocity is
$S=[1000-(200+200\left)\right]$
$=600\text{m}$ at $20m{s}^{-1}$
and time of travel for constant velocity is
$t_3=\frac{S}{v}=\frac{600}{20}=30s$

Hence, total time for travel
$=t_1+t_2+t_3=70s.$

It can be shown on $v-t$ graph as