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Q.

A particle of mass 0.5 kg is moving in xy plane with acceleration a→=3i^+2tj^m/s2. At time t, its velocity is given by  V→=2ti^+3t2j^m/s. Then rate of work done on the particle at t = 1 sec is

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a

6 watt

b

8 watt

c

12 watt

d

24 watt

answer is A.

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Detailed Solution

(1)Force acting on the particle,    F→=ma→=0.53i^+2tj^N∴  Power,   P=F→.V→=0.53i^+2tj^  .  2ti^+3tj^=12  6t+6t3  wattP=3t+t3  wattAt   t=1  sec,    P=31+1watt=6  watt.
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