Physics M.C.Q (1 Marks)

$\left(\frac{u}{3}\right)^2=u^2-2 a x$

$2 a x=u^2-\frac{u^2}{9}$

$2 a x=\frac{8 u^2}{9}.........(1)$

Similarly from starting

$v^2=u^2+2 ax$

$0= u ^2-2 ax _2$

$2 ax _2= u ^2..........(2)$

$By (1) /(2)$

$\frac{ x }{ x _2}=\frac{8}{9}$

$\frac{24}{ x _2}=\frac{8}{9}$

$x _2=27\,cm$

$-H =4 \times 4-\frac{1}{2} \times 10 \times 4^2$

$-H =16-80$

$-H =-64$

$H =64\,m$

$V =\frac{ dx }{ dt }=\tan \theta$

$\frac{ V _{1}}{ V _{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}=\frac{1}{\sqrt{3}}$

$=0+\frac{ a }{2}(2 n -1)$

$S _{r th } \propto(2 n -1)$

$\Rightarrow S_{1 s }, S _{2 ed }, S _{3 ed }, S _{\text {4th }}$

$=[2(1)-1]:[2(2)-1]:[2(3)-1]:[2(4)-1]$

$=1: 3: 5: 7$

$S = ut +\frac{1}{2} at ^{2}$

$1.5= u (0.1)+\frac{1}{2} \times 10(0.1)^{2}$

$1.5=(0.1) u +0.05$

$u =15-0.5$

$\quad=14.5 m / s$

$80^{2}=20^{2}+2 \times 10 h$

$h =300 m$

$\begin{array}{l}
\therefore \,Net\,velocity\,of\,preeti\,on\,moving\,escalator\,with\,respect\,\\
to\,the\,ground\,\\
\,\,\,\,\,\,\,\,\,\,\,v = {v_1} + {v_2} = \frac{d}{{{t_1}}} + \frac{d}{{{t_2}}} = d\left( {\frac{{{t_1} + {t_2}}}{{{t_t}{t_2}}}} \right)\\
The\,time taken by her to walk up on the moving escalatror will be\\
\,\,\,\,\,\,t = \frac{d}{v} = \frac{d}{{d\left( {\frac{{{t_1} + {t_2}}}{{{t_1}{t_2}}}} \right)}} = \frac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}
\end{array}$

$\frac{ ds }{ dt }=\alpha t +\beta t ^{2}$

$\int_{s_{1}}^{s_{2}} d s=\int_{1}^{2}\left(\alpha t+\beta t^{2}\right) d t$

$S_{2}-S_{1}=\left[\frac{\alpha t^{2}}{2}+\frac{\beta t^{3}}{3}\right]_{1}^{2}$

As particle is not changing direction So distance $=$ displacement.

Distance $=\left[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}\right]$

$=\frac{3 \alpha}{2}+\frac{7 \beta}{3}$

Average acceleration $=\frac{\text { Change in velocity }}{\text { Time }} \Rightarrow a_{ av }=\frac{\int \limits_{t_1}^{t_2} a d t}{t_2-t_1}$

The acceleration from zero to $5 \,s$ is $a=\frac{0-20}{5-0}=\frac{-20}{5}=-4 \,ms ^{-2}$

From $5 \,s$ to $10 \,s$

$a=\frac{20-0}{10-5}=4 \,ms ^{-2}$

$a=\frac{\text { Total change in velocity }}{\text { Time }}$

$=\frac{20-20}{10-0}=0 \,ms ^{-2}$

From $C$ to $B$ the time interval of travelling is same.

So, $v_{ av }=\frac{v_2+v_3}{2}=\frac{2+4}{2}=3 \,m / s$

Now, first half is covered with $6 \,ms ^{-1}$ and second half with $3 \,ms ^{-1}$. So when distances are same.

$v_{ av }=\frac{2 v_1 v_2}{v_1+v_2}=\frac{2 \times 6 \times 3}{6+3}=4 \,ms ^{-1}$

$v_{ av }=4 \,ms ^{-1}$

$ = u + \frac{{3{t^3}}}{3} + \frac{{2{t^2}}}{2} + 2t = u + {t^3} + {t^2} + 2t$

$ = 2 + 8 + 4 + 4 = 18\;m/s$     (As $t = 2\, sec$)

$\frac{{dx}}{{dt}} = \frac{d}{{dt}}(2 - 5t + 6{t^2}) = (0 - 5 + 12t)$

For initial velocity $t = 0$, hence $v = - 5\;m/s$.

At $t = 0,\,x = - 8 + 4a = 4a - 8$

$v = \frac{{dx}}{{dt}} = 4 + 2a(t - 2)$

At $t = 0,\;\;v = 4 - 4a = 4(1 - a)$

But acceleration,$a = \frac{{{d^2}x}}{{d{t^2}}} = 2a$

$ = \frac{1}{2} \times 1 \times 20 + (20 \times 1) + \frac{1}{2}(20 + 10) \times 1 + (10 \times 1)$

$ = 10 + 20 + 15 + 10 = 55\;m$

Region $AB$ shows that graph is parallel to time axis i.e. velocity is zero. Hence acceleration is zero.

Region $BC$ shows that graph is bending towards displacement axis i.e. acceleration is positive.

Region $CD$ shows that graph having constant slope i.e. velocity is constant. Hence acceleration is zero.

$t=0 \text { to } t=2 \quad \quad t=6 \text { to } t=8$

$v=20 \,m / s \quad \quad \quad v=-20 \,m / s$

$a_{\text {avg }}=\frac{\Delta v}{\Delta t}=\frac{-20-20}{8}=\frac{-40}{8}=-5\,ms ^{-2}$

$a_{\text {avg }}=-5 \,ms ^{-2}$

$ = \left( {\frac{1}{2} \times 2 \times 3.6} \right) + \left( {8 \times 3.6} \right) + \left( {\frac{1}{2} \times 2 \times 3.6} \right)$= $36m$

Displacement $x = \int {v\;dt = \int {a\omega } } \cos \omega t\;dt = a\sin \omega t$

$60t + 2.5 = 70t - \frac{1}{2} \times 20 \times {t^2}$

==> $10{t^2} - 10t + 2.5 = 0$==> ${t^2} - t + 0.25 = 0$

$\therefore $ $t = \frac{{1 \pm \sqrt {1 - 4 \times (0.25)} }}{2} = \frac{1}{2}hr$

Distance travelled $S =$ Area under curve $= 2+2=4m$

==> $\int\limits_{{v_0}}^0 {vdv = - \alpha \int\limits_0^S {{x^2}dx} } $ ==> $\left[ {\frac{{{v^2}}}{2}} \right]_{{v_0}}^0 = - \alpha \left[ {\frac{{{x^3}}}{3}} \right]_0^S$

==> $\frac{{v_0^2}}{2} = \frac{{\alpha \,{S^3}}}{3}$==> $S = {\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}$

$v=3 \alpha t^2+2 \beta t+\gamma$

$v_{t=0}=v_i=\gamma$

$a=6 \alpha t+2 \beta: a_{t=0}=a_i=2 \beta$

$\therefore \frac{v_i}{a_i}=\frac{\gamma}{2 \beta}$

at $t=0 ; y=0$

at $t=2, y=4$

$\frac{\mathrm{d}^{2} \mathrm{V}}{\mathrm{dt}^{2}}=12-36 \mathrm{t}>0 \Rightarrow$ for $\mathrm{t}=0$ so $\mathrm{v}_{\min }$ will be at

$t=0$

Equation of line $R S$ is $y=-m x+C$

Or $y=-\left(\frac{v_2}{v_1}\right) x+v_2 t$

or $\quad v_1 y=-v_2 x+v_1 v_2 t$

Equation of line $O P$ is

$y=x$

Point $P$ is the point of intersection, we get

$x_P=y_P=\frac{v_1 v_2 t}{v_1+v_2}$

$O P=\sqrt{x_P^2+y_P^2}$

$=\frac{\sqrt{2} v_1 v_2 t}{v_1+v_2}$

$v_3=\frac{O P}{t}=\frac{\sqrt{2} v_1 v_2}{v_1+v_2}$

$\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^{2}-12 \mathrm{t}+3$

$a=\frac{d v}{d t}=6 t-12$

$a=0$

$t=2 \sec$

Velocity at $t=2 \mathrm{sec}$

$\mathrm{v}=-9 \mathrm{m} / \mathrm{s}$

Comparing the given equation with general equation of displacement, $s=s_0+u t+\frac{1}{2} a t^2$, we get

$u=+20 \text { unit and } a=-10 \text { unit }$

$d v=a d t$

$\therefore \int \limits_2^v d v=\int \limits_0^2\left(3 t^2+2 t+2\right) d t$

$\text { or } v=18\,m / s$

$ = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$ $ = \frac{{\left[ {10 + 2{{(5)}^2}} \right] - \left[ {10 + 2{{(2)}^2}} \right]}}{3} = \frac{{60 - 18}}{3}$$ = 14\;m/{s^2}$.

For e.g. when an object is thrown vertically upward, at highest point velocity is zero but acceleration on it due to gravity is non-zero.

Option $D$ is correct as the derivative of velocity will be zero in case velocity is zero for a time interval,

$t=12 s$

to come in rest. ln this case by using

$v^{2}=u^{2}+2 a s$

$\Rightarrow 0=(40)^{2}+2 \mathrm{a} \times 1000$

$\Rightarrow \mathrm{a}=-0.8 \mathrm{m} / \mathrm{s}^{2}$

Distance travelled by body $\mathrm{A}$ in $5^{\text {th }}$ sec and distance travelled by body $\mathrm{B}$ in $3^{\mathrm{rd}}$ sec. of its motion are

equal.

$0+\frac{a_{1}}{2}(2 \times 5-1)=0+\frac{a_{2}}{2}[2 \times 3-1]$

$9 a_{1}=5 a_{2} \Rightarrow \frac{a_{1}}{a_{2}}=\frac{5}{9}$

$\frac{d x}{d t}=\frac{k}{b} e^{-b t}, \frac{d^{2} x}{d t^{2}}=-k e^{-b t}$