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)
B
m3P(v2v1)
C
3Pm(v22v21)
D
m3P(v22v21)

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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