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Question

A man is 45m behind the bus when the bus starts accelerating from rest with acceleration of 2.5ms-2. With what minimum velocity should the man start running to catch the bus?


A

12ms-1

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B

14ms-1

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C

15ms-1

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D

16ms-1

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Solution

The correct option is C

15ms-1


Step 1: Given Data:

Distance, s=45m

Acceleration of bus, a=2.5ms-2

u=0m/s speed of bus

Step 2: Formula Used:

s=ut+12at2

Where, u=initial speed

v=Final speed

a=acceleration

t=time

Step 3: Calculating the required speed:

s=0+12×2.5×t2 [Where, ithe nitial velocity is zero as bus starts from rest.]

s=2.52t2….(1)

From the above figure, man needs to cover distance (45+s)m at time t to catch the bus

ut=45+s

ut=45+2.52t2 [from(1)]

u=45t+2.52t ….(2)

For minimum velocity,

dudt=0

-45t2+1.25=0

45=1.25t2

t2=451.25=36

t=36s

t=6s

From equation(2),

umin=456+1.25×6=7.5+7.5=15m/s

Hence, option C is the correct answer.


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