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2nd Equation of Motion

A passenger i...

Question

A passenger is standing 10 $m$ away from a bus. The bus begins to move with constant acceleration of 5 $m / s^{2}$ away from passenger. With what minimum speed must passenger run to catch bus.

A

10

m / s

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B

8

m / s

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C

12

m / s

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D

5 \sqrt{2}

m / s

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Solution

The correct option is A 10 $m / s$

Let man catches bus after time t, and let the bus travels a distance d and acheives a speed v.
Then $v = 0 + 5 t$ [For bus]
$v = \frac{10 + d}{t}$ [For man] and $d = \frac{1}{2} \times 5 t^{2}$ [For bus]
Solving all we get is d = 10 m & v = 10 m/s & t = 2 sec

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