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Question

If the distance ‘s’ traveled by a particle in time t is s=asint+bcos2t, then the acceleration at t=0 is


A

a

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B

-a

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C

4b

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D

-4b

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Solution

The correct option is D

-4b


Step 1 Given data:

The distance ‘s’ traveled by a particle in time t is s=asint+bcos2t

Step 2 Formula used:

Acceleration(a) is the second derivative of displacement(s).

That is, a=d2sdt2

Step 3 Acceleration at t=0

Acceleration, a=d2sdt2....(1)

That is, differentiating the displacement with respect to time

dsdt=d(asint+bcos2t)dtdsdt=acost2b(sin2t)

Once again, differentiating the displacement with respect to time,

d2sdt2=d(acost2b(sin2t))dtd2sdt2=asint4bcos2t...(2)

So, from equation (1) and (2),

a=asint4bcos2t

Therefore, acceleration at t=0 is,

a=asin×04bcos2×0a=a×0-4b×1a=0-4ba=-4b

The distance ‘s’ traveled by a particle in time t is s=asint+bcos2t, then the acceleration at t=0 is -4b. Hence, option D is correct.


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