Question

1.22. The velocity of a particle moving in the positive direction
of the x axis varies as v = al/x, where a is a positive constant.
Assuming that at the moment t = 0 the particle was located at the
point x = 0, find:
(a) the time dependence of the velocity and the acceleration of the
particle;
(b) the mean velocity of the particle averaged over the time that
the particle takes to cover the first s metres of the path.

Answer 1

Topic :- Motion in straight line

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

Given that, the velocity of a particle moving in the positive direction of the x axis varies as :-

$\longrightarrow\:\:\sf v = a \sqrt{x}$

Velocity can be calculated by the given below formula :

$\longrightarrow\:\:\sf v = \dfrac{dx}{dt}$

$\longrightarrow\:\:\sf a \sqrt{x} = \dfrac{dx}{dt}$

$\longrightarrow\:\:\sf a.dt= \dfrac{dx}{ \sqrt{x} }$

By integrating both the sides we get :

$\longrightarrow \: \: \displaystyle \sf \int\limits_{0}^{t} a\, dt=\int\limits_{0}^{x} \dfrac{dx}{ \sqrt{x} }$

$\longrightarrow \: \: \displaystyle \sf a\int\limits_{0}^{t} dt=\int\limits_{0}^{x} \dfrac{dx}{ {x}^{ \frac{1}{2} } }$

$\longrightarrow \: \: \displaystyle \sf a\int\limits_{0}^{t} dt=\int\limits_{0}^{x} {x}^{ - \frac{1}{2}} \: dx$

$\longrightarrow \: \: \displaystyle \sf at=\int\limits_{0}^{x} {x}^{ - \frac{1}{2}} \: dx$

$\longrightarrow \: \: \displaystyle \sf at=\int\limits_{0}^{x} \dfrac{ {x}^{ - \frac{1}{2} + 1}}{ \dfrac{ - 1}{2} + 1 }$

$\longrightarrow \: \: \displaystyle \sf at=\int\limits_{0}^{x} \dfrac{ {x}^{ - \frac{1 + 2}{2}}}{ \dfrac{ - 1 + 2}{2} }$

$\longrightarrow \: \: \displaystyle \sf at=\int\limits_{0}^{x} \dfrac{ {x}^{ \frac{1}{2}}}{ \dfrac{ 1}{2} }$

$\longrightarrow \: \: \displaystyle \sf at=\int\limits_{0}^{x} 2{x}^{ \frac{1}{2}}$

$\longrightarrow \: \: \displaystyle \sf at=\Bigg[2{x}^{ \frac{1}{2}} \Bigg]^{x}_{0}$

$\longrightarrow \: \: \displaystyle \sf at=\Bigg[2{x}^{ \frac{1}{2}} - 2{(0)}^{ \frac{1}{2}} \Bigg]$

$\longrightarrow \: \: \displaystyle \sf at=\Bigg[2{x}^{ \frac{1}{2}} - 0\Bigg]$

$\longrightarrow \: \: \displaystyle \sf at = 2{x}^{ \frac{1}{2}}$

$\longrightarrow \: \: \displaystyle \sf at = 2 \sqrt{x}$

$\longrightarrow \: \: \displaystyle \sf \sqrt{x} = \dfrac{at}{2}$

Squaring both the sides we get :

$\longrightarrow \: \: \displaystyle \sf x = \dfrac{ {a}^{2} {t}^{2} }{4}$

The time dependence of the velocity and the acceleration of the particle :-

$\longrightarrow\:\:\sf v = \dfrac{dx}{dt}$

$\longrightarrow\:\:\sf v = \dfrac{d}{dt} \bigg( \dfrac{ {a}^{2} {t}^{2} }{4} \bigg )$

$\longrightarrow\:\: \underline{ \boxed{\sf v = \dfrac{ 2{a}^{2} t }{4}}}$

Now, the mean velocity of the particle averaged over the time that the particle takes to cover the first s metres of the path :

First we need to find the Time taken to cover first "s" distance :

$\dashrightarrow\:\:\sf Distance (x) = Distance (s) = \dfrac{a^2t^2}{4}$

$\dashrightarrow\:\:\sf s = \dfrac{a^2t^2}{4}$

$\dashrightarrow\:\:\sf 4s = a^2t^2$

$\dashrightarrow\:\:\sf t^2 = \dfrac{4s}{ {a}^{2} }$

$\dashrightarrow\:\:\sf t= \sqrt{\dfrac{4s}{ {a}^{2} }}$

Finding the mean velocity of the particle :-

$:\implies \sf Average \: Velocity = \dfrac{Total \: Displacement}{Total \: time \: taken}$

$:\implies \sf V_{av} = \dfrac{s}{t}$

$:\implies \sf V_{av} = \dfrac{s}{ \sqrt{ \dfrac{4s}{ {a}^{2} }} }$

$:\implies \sf V_{av} = \dfrac{ \sqrt{s} \times \sqrt{s} }{ \dfrac{2 \sqrt{s} }{ a } }$

$:\implies \sf V_{av} = \dfrac{ \sqrt{s} }{ \dfrac{2}{ a } }$

$:\implies \underline{ \boxed{ \sf V_{av} = \dfrac{ a\sqrt{s} }{ 2 } }}$

Answer 2

Topic :- motion in straight line

Given that the velocity of a particle moving in the positive direction the x axis varies

as:-