NEET Practice Questions (MCQs) with Answers & Solutions

(c)

Distance travelled by the particle in nth second is 

    $S_{nth}=u+\frac{1}{2}a\left(2n-1\right)$

where u is initial speed and a is acceleration of the particle.

Here, n=3, u=0, a=$\frac{4}{3}m/s^2$

$\therefore S_{3rd}=0+\frac{1}{2}\times \frac{4}{3}\times \left(2\times 3-1\right)$

                 = $\frac{4}{6}\times 5$

                =$\frac{10}{3}m$

Alternatively : Distance travelled in the 3rd second = distance travelled in 3s - distance travelled in 2s

As, u=0,

 $S_{\left(3rd s\right)}=\frac{1}{2}a.3^2-\frac{1}{2}a.2^2=\frac{1}{2}.a.5$

$Given a=\frac{4}{3}m{s}^{-2}$

  $\therefore S_{\left(3rd s\right)}=\frac{1}{2}\times \frac{4}{3}\times 5=\frac{10}{ 3}m$

The problem requires kinematics equations of motion. 

Let u and v  the first and final velocities of particle and a and s be the constant acceleration and distance covered by it. 

from third equation of motion 

        $$\begin{gathered} v^2=u^2+2as \\ \Rightarrow {\left(20\right)}^2={\left(10\right)}^2+ 2a\times 135 \\ or a=\frac{300}{2\times 135}=\frac{10}{9}m{s}^{-2} \end{gathered}$$

Now using first equation of motion,

                 v=u+ at

or     t=$\frac{v-u}{a}=\frac{20-10}{\left(10/9\right)}=\frac{10\times 9}{10}=9s$

$\mathrm{Average} \mathrm{velocity}=\frac{∆\mathrm{x}}{∆\mathrm{t}}=\frac{{\mathrm{x}}_2-{\mathrm{x}}_1}{{\mathrm{t}}_2-{\mathrm{t}}_1}=\frac{28-48}{4}=-5 \mathrm{m}/\mathrm{s}$

$$\begin{gathered} x=t^3-3t^2+3t+3 \\ \Rightarrow v=\frac{dx}{dt}=3t^2-6t+3 \\ When it stops, v=0 \\ \Rightarrow 3t^2-6t+3=0 \\ \Rightarrow t^2-2t+1=0 \\ \Rightarrow {(t-1)}^2=0\Rightarrow t=1 sec \\ So, the particle stops only once at t=1 sec \end{gathered}$$

(A)

$$\begin{gathered} \mathrm{x}=4\mathrm{a}\left[\mathrm{t}+\mathrm{a} \sin \left(\frac{\mathrm{t}}{\mathrm{a}}\right)\right] \\ \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}} \\ \mathrm{v}=4\mathrm{a}\left[1+\cos \left(\frac{\mathrm{t}}{\mathrm{a}}\right)\right] \\ \mathrm{For} \mathrm{velocity} \mathrm{to} \mathrm{be} \mathrm{zero}, \\ 1+\cos \left(\frac{\mathrm{t}}{\mathrm{a}}\right)=0\Rightarrow \mathrm{t}=\mathrm{\pi a} \\ \mathrm{Putting} \mathrm{in} \mathrm{x}, \mathrm{x}=4{\mathrm{\pi a}}^2 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}^2={\mathrm{t}}^2+1 \\ 2\mathrm{x}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)=2\mathrm{t} \\ \mathrm{v}=\frac{\mathrm{t}}{\mathrm{x}} \\ \mathrm{a}=\frac{\mathrm{x}-\mathrm{vt}}{{\mathrm{x}}^2} \\ =\frac{1}{\mathrm{x}}-\frac{{\mathrm{t}}^2}{{\mathrm{x}}^3} \\ \mathrm{At} \mathrm{t}=1 \sec \\ \mathrm{a}=\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^3} \end{gathered}$$

Given that $\frac{dv}{dt}=4-2v$

dv=(4-2v)dt    or    $\frac{dv}{4-2v}=dt$

or $dv={}_0{}^v\int \frac{dv}{4-2v}={}_0{}^t\int dt$           or      ${\left[\frac{{\log }_e(4-2v)}{-2}\right]}_0^v ={\left[t\right]}_0^t$

or  ${\log }_{\mathrm{e}}\left\{4-2\left.\mathrm{v}\right\}\right.-{\log }_{\mathrm{e}}\left(4\right)=-2\mathrm{t}$   or          ${\log }_e\left\{\frac{4}{4}\right.-\left.\frac{2v}{4}\right\}=-2t$

$\Rightarrow \frac{4-2v}{4}=e^{-2t} \Rightarrow \frac{4}{4}-\frac{2v}{4}=e^{-2t} \Rightarrow 1-\frac{2v}{4}=e^{-2t}$

$\Rightarrow \frac{2v}{4}=1-e^{-2t}$                     or        $=2(1-e^{-2t})$

Horizontal component of velocity remains constant during projectile motion

so, 

$$\begin{gathered} v\cos \alpha =u\cos \theta \\ \Rightarrow v=\frac{u\cos \theta }{\cos \alpha } \end{gathered}$$

Horizontal velocity is always constant during projectile motion, when only gravitational force act on the body.

So, $v \cos \alpha =u \cos \theta$

$v\times \frac{\sqrt{3}}{2}=20\sqrt{3}\times \frac{1}{2}\Rightarrow v=20 m/s$

 $$\begin{gathered} \mathrm{Let} \mathrm{time} \mathrm{to} \mathrm{reach} \mathrm{P} \mathrm{from} \mathrm{A} \mathrm{be} \mathrm{t} \\ \mathrm{H}=\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{t}=\sqrt{\frac{2\mathrm{H}}{\mathrm{g}}} \\ \mathrm{Time} \mathrm{to} \mathrm{reach} \mathrm{C} \mathrm{from} \mathrm{P} \mathrm{be} \mathrm{t}\text{'} \\ \frac{\mathrm{H}}{2}=\frac{1}{2}\mathrm{g}{\left(\mathrm{t}\text{'}\right)}^2 \\ \mathrm{t}\text{'}=\sqrt{\frac{\mathrm{H}}{\mathrm{g}}}= \sqrt{\frac{2\mathrm{H}}{2\mathrm{g}}}= \frac{\mathrm{t}}{\sqrt{2}} \\ \mathrm{So} \mathrm{total} \mathrm{time} \mathrm{to} \mathrm{go} \mathrm{from} \mathrm{A} \mathrm{to} \mathrm{C} \\ \mathrm{t}+\mathrm{t}\text{'}= \mathrm{t}+\frac{\mathrm{t}}{\sqrt{2}} \\ = \mathrm{t} \left[1+\frac{1}{\sqrt{2}}\right] \end{gathered}$$