Q. 83.8( 4 Votes )

# A bus begin to move with an acceleration of 1 m/s^{2}. 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 s

B. 5 s

C. 6 s

D. 7 s

Answer :

For bus:

Using the second equation of motion: s = ut + at^{2}

where: s_{1} = distance covered = ? m

u = initial velocity = 0 m/s

a = acceleration = 1 m/s^{2}

t = time = ts

s_{1} = ut + at^{2}

s_{1} = 0 × t + × 1 × t^{2} = t^{2}/2

Distance travelled by the man, s_{2} = 10t

s_{2} – s_{1} = 48 m

10t - t^{2}/2 = 48

20t – t^{2} = 96

20t – t^{2} – 96 = 0

t^{2} – 20t +96 =0

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

t = 12s or 8s

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