Question

.A bus starts from rest with an acceleration of 1m/s 2 . A man who is 48m behind the bus starts
with a uniform velocity of 10m/s. what will be the minimum time after which the man will
catch the bus.

Answer 1

$\huge{\underline{\tt{\green{Answer-}}}}$

$t=8s$

Explanation:

Let man will catch the bus after time t.Since the distance from man position is same for both at time t

Distance covered by bus = Distance covered by man -48

$\dfrac {1}{2}at^2 = vt -48$

$\dfrac {1}{2} (1) t^2 = 10t -48$

t^2 −20t+96=0

(t−12)(t−8)=0

t=8s (minimum value) or t=12s

t= 8 s

Answer 2

Answer:

8 s is answer of question