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Question

A particle of mass 2 kg starts moving in a straight line with an initial velocity of 2 m/s, at a constant acceleration of 2 m/s2. Then, rate of change of kinetic energy

A
is four times the velocity at any moment.
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B
is two times the displacement at any moment.
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C
is four times the rate of change of velocity at any moment.
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D
is constant throughout.
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Solution

The correct option is A is four times the velocity at any moment.
We know that, Kinetic energy K=12mv2
Differentiating on both sides with respect to time, we get
dKdt=mv.dvdt
dKdt=(mdvdt)v=(ma)v
From the data given in the question,
m=2 kg and a=2 m/s2
dKdt=4v .....(1)
Since the given acceleration is constant, we can use kinematic formulae
Using, v=u+at in (1), we get
dKdt=4u+4at=8(1+t)
So, option (d) is wrong.
From (1), we can conclude that option (c) is also wrong.
Using v2=u2+2as in (1) and substituting the values, we get
dKdt=81+2s
So, option (b) is also wrong.
Thus, option (a) is the correct answer.

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