Question

A particle moving in a straight line having acceleration which varies with its velocity as a = -kv^n. Here k is a
constant. For what value of n the average value of velocity of the particle averaged over the time, till it stops,
is one third the initial velocity. (n 1 or 2)
​

Answer 1

Answer:

n=1/2.

Explanation:

1) Let the initial velocity of particle be $v_0$

Then, particle stops in time 't'.

$a=-kv^n\\ \\=>\frac{dv}{dt}=-kv^n\\ \\=>dv*v^{-n}=-kdt\\ \\=>\int\limits^{0}_{v_o} {v^{-n}} \, dv=-k\int\limits^t_0  \, dt \\ \\=>\frac{v_0^{1-n}}{1-n}=kt$

2) Now,

Here Limits are same as above.

$v_{avg}=\frac{\int_{v_o}^{0} vdt}{\int_0^tdt} =\frac{\int v* \frac{dv}{dv/dt}}{\int dt}\\ \\=\frac{\int v* \frac{dv}{a}}{\int dt}\\ \\=\frac{\int v* \frac{dv}{-kv^{n}}}{\int dt}\\ \\=\frac{\int v^{1-n}dv }{-kt} =\frac{v_0^{2-n}}{kt*(2-n)} =\frac{v_0^{2-n}}{\frac{v_0^{1-n}}{1-n}*(2-n)}=\frac{v_0*(1-n)}{(2-n)}$

3) it is given that,

$v_{avg}=\frac{v_0}{3}\\ \\=> \frac{v_0*(1-n)}{(2-n)}=\frac{v_0}{3}\\ \\=>\frac{1-n}{2-n}=\frac{1}{3}\\ \\=>n=\frac{1}{2}$

Hence, Required answer is

$\boxed{n=\frac{1}{2}}$

Answer 2

Answer:

n=1/2

Explanation:

the correct is n=1/2