Question

A car and a truck move in the same straight line at the same instant of time from the same point. The car moves with a constant velocity of $40\text{m/s}$ and truck starts with a constant acceleration of $4{\text{m/s}}^2$. Find the time $t$ that elapses before the truck catches with the car and also find the greatest distance $S_{max}$ between them prior to it

• A. $t=5\text{sec},S_{max}=100\text{m}$
• B. $t=20\text{sec},S_{max}=200\text{m}$
• C. $t=15\text{sec},S_{max}=50\text{m}$
• D. $t=8\text{sec},S_{max}=100\text{m}$

Answer 1

The correct option is B $t=20\text{sec},S_{max}=200\text{m}$

Velocity of car, $u=40\text{m/s}$,

Acceleration of the truck, $a_t=4{\text{m/s}}^2$

The distance travelled by car in time $t$, $s_1=ut=40t$

The distance travelled by truck in the same time $t$,

$s_2=\frac{1}{2}\left(a_t\right)t^2=\frac{1}{2}\times 4\times t^2=2t^2$

The separation between them at any time $t$,

$s=s_1-s_2=40t-2t^2...\left(1\right)$

When truck catches the car, $s=0$

$\Rightarrow s_1-s_2=0$

$\Rightarrow 40t-2t^2=0$

$\Rightarrow 2t(20-t)=0$

$t=20\text{sec}$ or $t=0\text{sec}$

At $t=0$, both start moving from same point.
Hence, at $t=20\text{sec}$, the second body will catch first body.

For $s$ to be minimum or maximum,
$\frac{ds}{dt}=0,$

$\therefore \frac{d(40t-2t^2)}{dt}=0,$

$\Rightarrow 40-4t=0$

$\therefore t=10\text{sec}$

Now check $s$ is minimum or maximum by doing double derivative test.

Differentiating equation $\left(1\right)$ again

$\frac{d^{2}s}{d{t}^2}=-4$

The double derivative of $s$ w.r.t $t$ is negative.

Hence, at $t=10\text{sec}$, distance between them will be maximum, which is given as
$S_{max}=40\left(10\right)-2(10{)}^2=200\text{m}$

Hence, option (b) is the correct answer.