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Question

A particle is moving on a straight line, where its position is a function of time t given by s=at2+bt+60. If it is known that the particle comes to rest after 4seconds at a distance of 16metres from the starting position (t=0), then the retardation in its motion is


A

-1m/s2

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B

54m/s2

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C

12m/s2

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D

-54m/s2

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Solution

The correct option is B

54m/s2


Step 1: Given data

Distance, s=at2+bt+6

Step 2: Find velocity, v

Velocity is the rate of change of displacement with respect to the time.

s=at2+bt+6

Differentiating with respect to the time

v=dsdtv=2at+beq1

Step 3: Find acceleration, A

After 4seconds,v=0 and distance s=16m

Putting these value in the above equation, we get the value of a

0=2a×4+b8a+b=0s=at2+bt+616=16a+4b+616=16a+4(-8a)+6a=58

Acceleration is the rate of change of the velocity with respect to the time.

A=dvdtA=ddt2at+bA=2aA=2×-58A=-54ms-2

Retardation =54ms-2

Hence, option B is correct.


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