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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)=46t2. 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/s2
Given the variation of the speed of the particle is:
v(t)=46t2
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)=464=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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