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. $\frac{3P}{m}(v_2^2-v_1^2)$
• B. $\frac{m}{3P}(v_2-v_1)$
• C. $\frac{m}{3P}(v_2^3-v_1^3)$
• D. $\frac{m}{3P}(v_2^2-v_1^2)$

Answer 1

The correct option is C $\frac{m}{3P}(v_2^3-v_1^3)$

$P=FV$

$P=mav$

$P=mv.\frac{dv}{ds}.v$

$\frac{P}{m}.ds=v^2.dv$

Integrate both sides under limits $Vfrom{v}_{1}to{v}_2$

$dsfrom0toS$

We get $\left(\frac{v^3}{3}\right)=\frac{P}{m.s}$

$\frac{1}{3}({v_2}^3-{v_1}^3)=\frac{P}{m.s}$

$\frac{M}{3}P({v_2}^3-{v_1}^3)=S$