Question

Particle located at x=0 at time t=0
Start moving along positive
x direction with Veloty v which
Varried as V=kx1/2
find Velocity
varies with Time
​

Answer 1

Given,

$\longrightarrow\sf{v=k\sqrt x}$

where $\sf{k}$ is a constant.

We have to find velocity in terms of time.

We can differentiate our equation with respect to $\sf{x}$ as,

$\longrightarrow\sf{\dfrac{dv}{dx}=\dfrac{d}{dx}\left[k\sqrt x\right]}$

$\longrightarrow\sf{\dfrac{\left(\dfrac{dv}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=k\cdot\dfrac{d}{dx}\left[\sqrt x\right]}$

$\longrightarrow\sf{\dfrac{a}{v}=\dfrac{k}{2\sqrt x}}$

$\longrightarrow\sf{\dfrac{a}{k\sqrt x}=\dfrac{k}{2\sqrt x}}$

$\longrightarrow\sf{a=\dfrac{k^2}{2}}$

$\longrightarrow\sf{\dfrac{dv}{dt}=\dfrac{k^2}{2}}$

$\longrightarrow\sf{dv=\dfrac{k^2}{2}\ dt}$

Integrating for a velocity from $\sf{v=u}$ to $\sf{v=v}$ and for a time from $\sf{t=0}$ to $\sf{t=t.}$

$\displaystyle\longrightarrow\sf{\int\limits_u^v dv=\int\limits_0^t\dfrac{k^2}{2}\ dt}$

$\displaystyle\longrightarrow\sf{\big[v\big]_u^v=\dfrac{k^2}{2}\big[t\big]_0^t}$

$\displaystyle\longrightarrow\sf{v-u=\dfrac{k^2}{2}(t-0)}$

$\displaystyle\longrightarrow\underline{\underline{\sf{v=u+\dfrac{1}{2}\,k^2t}}}$

If $\sf{u=0,}$

$\displaystyle\longrightarrow\underline{\underline{\sf{v=\dfrac{1}{2}\,k^2t}}}$