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 body start from rest with a constant acceleration of $40m{s}^2$. The time that elapse before the second catches the first body is (in seconds)

Answer 1

The distance travelled by first body in time t, $s_1=40t$ .
The distance travelled by second body in time t,
$s_2=\frac{4t^2}{2}=2t^2$.
When second body catches first body $s_1-s_2=0$.
$\Rightarrow s_1-s_2=40t-2t^2=0$.
which gives t=0,20 s. At begining t=0 s . So they will again meet at t=20 s.