Question

Try this question many didn't get the ans..ans is opt3 explain how

Answer 1

Given that the particle is moving under constant power.

We know power $\sf{c}$ is the product of force acting on it and its velocity.

$\longrightarrow\sf{c=Fv}$

$\longrightarrow\sf{F=\dfrac{c}{v}}$

We know that $\sf{F=ma.}$ Then,

$\longrightarrow\sf{ma=\dfrac{c}{v}}$

$\longrightarrow\sf{a=\dfrac{c}{mv}}$

Acceleration is the first derivative of velocity wrt time.

$\longrightarrow\sf{\dfrac{dv}{dt}=\dfrac{c}{mv}}$

$\longrightarrow\sf{v\ dv=\dfrac{c}{m}\ dt}$

$\displaystyle\longrightarrow\sf{\int v\ dv=\int \dfrac{c}{m}\ dt}$

Since P and m are constant wrt time,

$\displaystyle\longrightarrow\sf{\dfrac{v^2}{2}=\dfrac{ct}{m}}$

$\displaystyle\longrightarrow\sf{v=\left[\dfrac{2ct}{m}\right]^{\frac{1}{2}}}$

Well, velocity is the first derivative of displacement wrt time.

$\displaystyle\longrightarrow\sf{\dfrac{dx}{dt}=\left[\dfrac{2ct}{m}\right]^{\frac{1}{2}}}$

$\displaystyle\longrightarrow\sf{dx=\left[\dfrac{2ct}{m}\right]^{\frac{1}{2}}\ dt}$

$\displaystyle\longrightarrow\sf{x=\int\left[\dfrac{2ct}{m}\right]^{\frac{1}{2}}\ dt}$

$\displaystyle\longrightarrow\sf{x=\sqrt{\dfrac{2c}{m}}\int t^{\frac{1}{2}}\ dt}$

Since $\displaystyle\sf{\int x^n\ dx=\dfrac{x^{n+1}}{n+1},\ n\neq-1,}$

$\displaystyle\longrightarrow\sf{x=\sqrt{\dfrac{2c}{m}}\cdot\dfrac{t^{\frac{3}{2}}}{\left(\dfrac{3}{2}\right)}}$

$\displaystyle\longrightarrow\sf{x=\sqrt{\dfrac{2c}{m}}\cdot\dfrac{2t^{\frac{3}{2}}}{3}}$

$\displaystyle\longrightarrow\sf{x=\dfrac{2}{3}\sqrt{\dfrac{2\cdot3c}{3m}}\,t^{\frac{3}{2}}}$

$\displaystyle\longrightarrow\sf{x=\dfrac{2}{3}\sqrt{\dfrac{2}{3}}\cdot\sqrt{\dfrac{3c}{m}}\,t^{\frac{3}{2}}}$

$\displaystyle\longrightarrow\sf{x=\left(\dfrac{2}{3}\right)^{1+\frac{1}{2}}\left(\dfrac{3c}{m}\right)^{\frac{1}{2}}\,t^{\frac{3}{2}}}$

$\displaystyle\longrightarrow\sf{x=\left(\dfrac{2}{3}\right)^{\frac{3}{2}}\left(\dfrac{3c}{m}\right)^{\frac{1}{3}\cdot\frac{3}{2}}\,t^{\frac{3}{2}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{x=\left[\dfrac{2}{3}\left(\dfrac{3c}{m}\right)^{\frac{1}{3}}\,t\right]^{\frac{3}{2}}}}}$

Hence 3rd option is the answer.