Question

A bus starts to move with an accleration of 1 m/s a man who is 48 m behind the bus runs with constant velocity of 10 m/s to catch it in how much time he will catch the bus.

Answer 1

 Hi,

Let after time t man caches the bus.

After time t bus has covered the distance =1/2 (2)t2 = t2/2
Therefore the total distance man has cover 48 = t2/2

Therefore,
vt=s

10t = 48+ t2/2
20t = 96+ t2
=> t2-20t + 96=0
=> (t-12) (t-8) = 0
=> t=12 or t=8

Therefore, minimun time in which man caches the bus = 8 sec.

Hope this helps,
and t2= 2square