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Question: A constant power P is applied to particle of mass m. The distance travelled by the particle when its...

A constant power P is applied to particle of mass m. The distance travelled by the particle when its velocity increases from v1 to v2 is : (Neglect friction) :

A

3Pm(v22v12)\frac{3P}{m}(v_{2}^{2} - v_{1}^{2})

B

m3P(v2v1)\frac{m}{3P}(v_{2} - v_{1})

C

m3P(v23v13)\frac{m}{3P}(v_{2}^{3} - v_{1}^{3})

D

m3P(v22v12)\frac{m}{3P}(v_{2}^{2} - v_{1}^{2})

Answer

m3P(v23v13)\frac{m}{3P}(v_{2}^{3} - v_{1}^{3})

Explanation

Solution

P = Fv or P = mav or p = m (vdvdx)\left( \frac{vdv}{dx} \right)v

or P = mv2 dvdx\frac{dv}{dx} or dx = mP\frac{m}{P}v2 dv

or 0xdx=mpv1v2v2dv\int_{0}^{x}{dx = \frac{m}{p}\int_{v_{1}}^{v_{2}}{v^{2}dv}}

or x = mP(v23v13)3\frac{m}{P}\frac{(v_{2}^{3} - v_{1}^{3})}{3}

or x = m3P(v23v13)\frac{m}{3P}(v_{2}^{3} - v_{1}^{3})