# A bus begin to move with an acceleration of 1 m/s2. A man who is 48 m behind the bus starts running at 10 m/s to catch the bus. The man will be able to catch the bus after.A. 8 sB. 5 sC. 6 sD. 7 s

For bus:

Using the second equation of motion: s = ut + at2

where: s1 = distance covered = ? m
u = initial velocity = 0 m/s
a = acceleration = 1 m/s2
t = time = ts

s1 = ut + at2

s1 = 0 × t + × 1 × t2 = t2/2

Distance travelled by the man, s2 = 10t

s2 – s1 = 48 m

10t - t2/2 = 48

20t – t2 = 96

20t – t2 – 96 = 0

t2 – 20t +96 =0

(t-12)(t-8) = 0

t = 12s or 8s

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