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Question

# A particle is moving in a circle of radius r=2 m. Its speed with time varies according to the relation v(t)=4−6t2. What is the magnitude of the total acceleration of the particle at t=2 sec ? Use the value of √769=27.73 for calculation.

A
2.11 m/s2
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B
3.46 m/s2
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C
3.1 m/s2
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D
1.5 m/s2
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Solution

## The correct option is B 3.46 m/s2Given the variation of the speed of the particle is: v(t)=4−6t2 Diffrentiating both sides w.r.t time, we get: dvdt=0+12t3 Now, tangential acceleration at is the rate of change of speed with time, i.e at=dvdt=12t3 At t=2 s, at=128=32 m/s2 Now speed at t=2 s: v(t=2)=4−64=52 m/s So, centripetal acceleration at t=2 sec ∴ac=v2r=(254)2=258 m/s2 Total acceleration aT of particle is aT=√a2c+a2t=√62564+94=√7698 =27.738=3.46 m/s2

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