Question

A bus starts from rest at constant acceleration 1 ms-2 a boy standing 48 m behind the bus start moving towards with constant speed 10 ms speed of the bus when it overtake boy

Answer 1

Suppose boy take time t to overtake the bus.

Now in time t distance travelled by bus would be equal to distance travelled by boy +48 m.

So distance covered by bus = ut + 1/2 *a*t*t
where u is initial velocity of bus and it is 0 since bus starts from rest.

so distance = 0+ 1/2 * 1 * t * t
= 0.5 *t * t meter

And distance travelled by boy = 10 * t meter

Hence
=> 10 *t +48 = 1/2 * t * t
=> t * t -20 *t -96 =0
=> t * t -24 * t + 4 *t -96=0
=> t *( t -24 ) + 4 *(t -24 )=0
=> ( t -24 )*(t + 4) =0
since t can not be negative so t =24 second.
So time taken to overtake bus = 24 second.

Hence speed of bus by first equation of motion is
v = u + a * t
v = 0+ 1 * 24
v =24 m/s