Question

A bus starts from rest with an acceleration of 1 m/s2. A
man who is 48 m behind the bus starts with a uniform
velocity of 10 m/s. Then the minimum time after which the
man will catch the bus :​

Answer 1

Explanation:

acceleration of the bus = 1m/sec^2

s = 48m

time taken by man = time taken by bus = x

acceleration of man = acceleration of the bus = 1

u of man = 0

v of man = 10m/s

t = (v - u)/a

t = 10s

Answer 2

Answer: 8 second

Explanation:Let the man catches the bus in t sec.

Before catching the bus=10tm

By applying 2nd eq.of motion

1÷2at^2+48=10t

1*1t^2-20t+96=0...........(1)

t^2-20t+96=0

By using splitting term method in eq (1)

(t-8)(t-12)

Hence the answer will be 8 seconds