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Solution :

When the second body catches the first, the distance travelled by each is the same. <br> `therefore 40t=1/2(4)t^2` or t=20S <br> Now, the distance s between the two bodies at any time t is `s=ut-1/2 at^2` <br> For s to be maximum, `(ds)/(dt)=0` or u-at=0 <br> or `t=u/a=40/4=10s` <br> Maximum Distance <br> `=40xx10-1/2xx4xx(10)^2=400-200`=200m