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)

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

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

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

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

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Solution

The correct option is A $\frac{m}{3 P} (v_{2}^{3} - v_{1}^{3})$
Constant power $P = F v = m a v$
So, we get $a = \frac{P}{m v}$
or $v \frac{d v}{d s} = \frac{P}{m v}$
or $v^{2} d v = \frac{P}{m} d s$
Integrating on both sides, we get
$\frac{P}{m} \int_{0}^{s} d s = \int_{v_{1}}^{v_{2}} v^{2} d v$
or
or $\frac{P}{m} s = \frac{1}{3} (v_{2}^{3} - v_{1}^{3})$
Thus we get $s = \frac{m}{3 P} (v_{2}^{3} - v_{1}^{3})$

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

The correct option is A m3P(v32−v31)Constant power P=Fv=mavSo, we get a=Pmvor vdvds=Pmvor v2dv=PmdsIntegrating on both sides, we getPm∫s0ds=∫v2v1v2dv or or Pms=13(v32−v31)Thus we get s=m3P(v32−v31)

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)