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Question

A particle moves along a straight line according to the law s=162t+3t3, where s is the distance of the particle from a fixed point at the end of tsec. The acceleration of the particle at the end of 2s is


A

36m/s2

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B

34m/s2

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C

36m

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D

None of these

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Solution

The correct option is A

36m/s2


Step1: Given data

Distance, s=162t+3t3

Time, t=2s

Step 2: Find Acceleration, a

Acceleration is the rate of change of the velocity, v with respect to the time and the velocity is the rate of the change of the distance or displacement with respect to the time.

s=16-2t+3t3v=dsdtv=-2+9t2a=dvdta=ddt-2+9t2a=18t

At,

t=2s,a=18×2a=36ms-2

Hence, option A is correct.


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