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 `40ms^-1`, and the second starts from rest with a constant acceleration of `4ms^-2`.Find the time that elapses before the second catches the first body. Find the also the greatest distance between them prior to it and time at which this occurs.

Explanation

The distance travelled by first body in time t, `s_1=40t`
The distance travelled by second body in time t,
`s_2=1/2xx4xxt^2=2t^2`
The separation between them at any time `t`
`s=s_1-s_2=40t=2t^2` …(i)
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`.
`(ds)/(dt)=(d)/(dt)(40t-2t^2)=40-4t` ...(ii)
If `(ds)/(dt)=0`, then `(d)/(dt)(40t-2t^2)=0`
or `40-2xx2t=0` or `t=10s`
Now check s is minimum or maximum doing double derivative test. Differentiating equation (ii) again `(d^2s)/(dt^2)=-4`
The double derivative of s w.r.t t is negative. Hence, we will get maximum value of the funciton at `t=10s`.
The greatest distance travelled them putting `t=10s` in Eq. (i),
we get `s=40xx10-2xx10^2=200m`