Question

plz solve 2nd question ​

Answer 1

Given that the velocity of the particle varies with displacement as,

$\sf{\longrightarrow v=\sqrt{9+4x}}$

$\sf{\longrightarrow v^2=9+4x\quad\quad\dots(1)}$

$\sf{\longrightarrow x=\dfrac{v^2-9}{4}}$

Differentiating wrt $\sf{v,}$

$\sf{\longrightarrow\dfrac{dx}{dv}=\dfrac{v}{2 }}$

or,

$\sf{\longrightarrow\dfrac{\left(\dfrac{dx}{dt}\right)}{\left(\dfrac{dv}{dt}\right)}=\dfrac{v}{2}}$

$\sf{\longrightarrow\dfrac{v}{a}=\dfrac{v}{2}}$

$\sf{\longrightarrow a=2\ m\,s^{-2} }$

Note that acceleration is independent of time. So we can make use of kinematic equations here.

By third equation of motion,

$\sf{\longrightarrow v^2=u^2+2ax}$

$\sf{\longrightarrow v^2=u^2+4x\quad\quad\dots(2)}$

Comparing (1) and (2) we get,

$\sf{\longrightarrow u=3\ m\,s^{-1}}$

Then by second equation of motion,

$\sf{\longrightarrow x=ut+\dfrac{1}{2}\,at^2}$

$\sf{\longrightarrow x=t^2+3t }$

Differentiating wrt $\sf{t}$ we get,

$\sf{\longrightarrow dx=(2t+3)\ dt\quad\quad\dots(3) }$

Hence the work done by all forces on the particle will be,

$\displaystyle\sf{\longrightarrow W=\int\limits_0^xF\ dx}$

$\displaystyle\sf{\longrightarrow W=\int\limits_0^xma\ dx}$

Taking $\sf{m=2\ kg}$ and $\sf{a=2\ m\,s^{-2},}$

$\displaystyle\sf{\longrightarrow W=4\int\limits_0^xdx}$

From (3),

$\displaystyle\sf{\longrightarrow W=4\int\limits_0^t(2t+3)\ dt}$

$\displaystyle\sf{\longrightarrow W=4\left[\left[t^2\right]_0^t+3\big[t\big]_0^t\right]}$

$\displaystyle\sf{\longrightarrow W=4t^2+12t}$

(i) At $\sf{t=2\ s,}$

$\displaystyle\sf{\longrightarrow W=4(2)^2+12(2)}$

$\displaystyle\sf{\longrightarrow\underline{\underline{W=40\ J}}}$

Hence (d) is the answer.

(ii) Average power is given by,

$\sf{\longrightarrow P_{av}=\dfrac{W}{t}}$

$\sf{\longrightarrow P_{av}=\dfrac{4t^2+12t}{t} }$

$\sf{\longrightarrow P_{av}=4t+12}$

Instαntaneous power is given by,

$\sf{\longrightarrow P_{in}=\dfrac{dW}{dt}}$

$\sf{\longrightarrow P_{in}=\dfrac{d}{dt}\left[4t^2+12t\right]}$

$\sf{\longrightarrow P_{in}=8t+12}$

We need to find the time at which,

$\sf{\longrightarrow P_{av}=\dfrac{3}{4}\,P_{in}}$

$\sf{\longrightarrow 4t+12=\dfrac{3}{4}\left(8t+12\right)}$

$\sf{\longrightarrow 4t+12=6t+9}$

$\sf{\longrightarrow\underline{\underline{t=1.5\ s}}}$

Hence (a) is the answer.