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Question

A constant power $$P$$ is applied to a particle of mass m. The distance travelled by the particle when its velocity increases from $$v_1$$ to $$v_2$$ is (neglect friction)


A
m3P(v32v31)
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B
m3P(v2v1)
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C
3Pm(v22v21)
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D
m3P(v22v21)
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Solution

The correct option is A $$\dfrac{m}{3P} (v_2^3 - v_1^3)$$
Constant power  $$P = Fv = mav$$
So, we get  $$a = \dfrac{P}{mv}$$
or $$v \dfrac{dv}{ds} = \dfrac{P}{mv}$$
or $$v^2 dv = \dfrac{P}{m} ds$$
Integrating on both sides, we get
$$\dfrac{P}{m}\displaystyle  \int_0^s  ds = \int_{v_1}^{v_2} v^2 dv$$ 
or  
or $$\dfrac{P}{m} s = \dfrac{1}{3} (v_2^3 - v_1^3)$$
Thus we get  $$s = \dfrac{m}{3P}  (v_2^3 - v_1^3)$$

Physics

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