Question

Two bodies start moving in the same straight line at the same instant of time from the same origin. The first body moves with a constant velocity of $40m{s}^{-1}$, and the second starts from rest with a constant acceleration of $4m{s}^{-2}$. Find the time that elapses before the second catches the first body. Find also the greatest distance between them prior to it and time at which this occurs.

Answer 1

The distance travelled by first body in time $t,s_1=40t$
The distance travelled by second body in time $t$,
$s_2=\frac{1}{2}\times 4\times t^2=2t^2$
The separation between them at any time $t$
$s=s_1-s_2=40t-2t^2.....\left(i\right)$
When second body catches first body, $s_1-s_2=0$
$s_1-s_2=40t-2t^2=0$
Which gives $t=0,20s$
At $t=0$, both start moving from same point. Hence, at $t=20s$, the second body will catch first body.
For $s$ to be minimum or maximum, $ds/dt=0$.
$\frac{ds}{dt}=\frac{d}{dt}(40t-2t^2)=40-4t.....\left(ii\right)$
If $\frac{ds}{dt}=0$, then $\frac{d}{dt}(40t-2t^2)=0$
or $40-2\times 2t=0$ or $t=10s$
Now check $s$ is minimum or maximum doing double derivative test. Differentiating equation (ii) again $\frac{d^{2}s}{d{t}^2}=-4$
The double derivative of $s$ w.r.t $t$ is negative. Hence, we will get maximum value of the function at $t=10s$.
The greatest distance travelled them putting $t=10s$ in Eq. (i), we get $s=40\times 10-2\times {10}^2=200m$.