A man is standing 40 m behind the bus. Bus starts with 1 $m / s e c^{2}$ constant acceleration and also at the same instant the man starts moving with constant speed 9 m/s. Find the time taken by man to catch the bus.

6 s

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10 s

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4 s

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13 s

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Solution

The correct option is B 10 s
Let after time 't'man will catch the bus
For bus
$x = x_{0} + u t + \frac{1}{2} a t^{2} x = 40 + 0 (t) + \frac{1}{2} (1) t^{2}$
$x = 40 + \frac{t^{2}}{2}$ ..................(1)
For man, x = 9t ................(2)
From (1) and (2)
$40 + \frac{t^{2}}{2} = 9 t t = 8 s o r t = 10 s .$

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A man is standing 40 m behind the bus. Bus starts with 1 m/sec2 constant acceleration and also at the same instant the man starts moving with constant speed 9 m/s. Find the time taken by man to catch the bus.

The correct option is B 10 sLet after time 't'man will catch the busFor busx=x0+ut+12at2x=40+0(t)+12(1)t2x=40+t22 ..................(1)For man, x = 9t ................(2)From (1) and (2)40+t22=9tt=8sort=10s.

A man is standing 40 m behind the bus. Bus starts with 1 $m / s e c^{2}$ constant acceleration and also at the same instant the man starts moving with constant speed 9 m/s. Find the time taken by man to catch the bus.