Question

A particle retards from a velocity $v_0$ while moving in a straight line. If the magnitude of deceleration is directly proportional to the square root of the speed of the particle, find its average velocity for the total time of its motion.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Method 1

So if deceleration is proportional to $\sqrt{v}$

So d = K $\sqrt{v}$

Acceleration = - deceleration

So a = - k $\sqrt{v}$

Now

$\frac{dv}{dt}=-K\sqrt{v}$

${\int }_{v_0}^v\frac{dv}{\sqrt{v}}{\int }_0^t-kdt$

∵ initial velocity of particle is $v_0$

$\Rightarrow 2\sqrt{v}{]}_{v_0}^v=-kt{]}_0^t$

2 ( $\sqrt{v}-\sqrt{v_0})$ = - kt

$\sqrt{v}=\sqrt{v_0}-\frac{k}{2}t$ ....... (i)

v = $v_0+\frac{k^2}{4}{t}^2-kt\sqrt{v_0}$ ..........(ii)

Particle starts motion at t=0 with velocity $v_0$ at t=T particle comes to rest as the particle is decelerating

$\Rightarrow$

Equation (i) 0 = $\sqrt{v_0}-\frac{k}{2}T$

$T=\frac{2\sqrt{v_0}}{k}$ ......... (iii)

Equation (ii) is velocity as function of t

$v_{avg}=\frac{totaldisplacement}{totaltime}=\frac{{\int }_0^{T}vdt}{{\int }_0^{T}dt}$

= $\frac{{\int }_0^T(v_0+\frac{k^2}{4}{t}^2-kt\sqrt{v_0})dt}{{\int }_0^{T}dt}=\frac{v_{0}t+\frac{k^2}{3}\frac{t^3}{3}-k\sqrt{v_0}\frac{t^2}{2}{]}_0^T}{t{]}_0^T}$

$v_{arg}=\frac{v_{0}T+\frac{k^2}{12}{T}^3-\frac{k\sqrt{v_0}}{2}{T}^2}{T}$

$=v.+\frac{k^2}{12}{T}^2-k\frac{\sqrt{v_0}}{2}T$

Putting value of T from equation (iii)

We get

$v_{arg}=\frac{v_0}{3}$

Method - 2

a = - k$\sqrt{v}$

$v=\frac{dv}{ds}=-k\sqrt{v}$

${\int }_{v_0}^0\sqrt{v}dv={\int }_0^s-kds$

$v_0-\frac{2(v_0{)}^{\frac{3}{2}}}{3}=-ks$

$s=\frac{2(v_0{)}^{\frac{3}{2}}}{3k}$ ........ (iv)

$v_{avg.}=\frac{Totaldisplacement}{Totaltime}=\frac{eq.\left(iv\right)}{eq.\left(iii\right)}=\frac{s}{t}$

= $\frac{\frac{2(v_0{)}^{\frac{3}{2}}}{3k}}{\frac{2\sqrt{v}}{k}}=\frac{v_0}{3}$

$v_{avg.}=\frac{v_0}{3}$