Motion In A Straight Line Practice Questions

${u}=\\sqrt{2 {gh}}$ Now, ${S}=\frac{{h}}{3} \;\;\;\;\;{a}=-g$ ${S}={ut}+\frac{1}{2} {at}^{2}$ $\frac{{h}}{3}=\\sqrt{2 {gh}} {t}+\frac{1}{2}(-{g}) {t}^{2}$ $t^{2}(\frac{g}{2})-\\sqrt{2 g h} t+\frac{h}{3}=0$ From quadratic equation $t_{1}, t_{2}={\\sqrt{2 g h} \±\\sqrt{2 g h-\frac{4 g}{2} \frac{h}{3}}}/{g}$ $\frac{{t}_{1}}{{t}_{2}}={\\sqrt{2 {gh}}-\\sqrt{\frac{4 {gh}}{3}}}/{\\sqrt{2 {gh}}+\\sqrt{\frac{4 {gh}}{3}}}$ $=\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}$

Consider 1st case, A car starts from rest at a constant acceleration $\alpha$. It means Acceleration, $a_{1}=\alpha$ initial velocity, $u=0$ $v=u+a_{1} t \Rightarrow v_{0}=0+\alpha t_{1}$ $\Rightarrow \;\; v_{0}=\alpha t_{1}$...(i) Consider 2nd case, The same car decelerates after some time at a constant acceleration $\beta$ and comes to rest. It means Acceleration, $a_{2}=\beta$ Final velocity $=0$ $\Rightarrow \;\; v_{0}=\beta t_{2}$...(ii) According to question, $t_{1}+t_{2}=t$ $\Rightarrow \;\; v_{0}(\;\frac{1}{\alpha }+\;\frac{1}{\beta })=t$ [from Eqs. (i) & (ii)] $\Rightarrow \;\; v_{0}=\;\frac{\alpha \beta t}{\alpha +\beta }$...(iii) If we draw the velocity-time graph for the given situation, it will be as follows$\therefore$ Distance travelled $=$ Area of $v-t$ graph $=\;\frac{1}{2} \times t \times v_{0}=\;\frac{1}{2} \times t \times \;\frac{\alpha \beta t}{(\alpha +\beta )}$ [from Eq. (iii)] $=\;\frac{1}{2} \;\frac{\alpha \beta t^{2}}{(\alpha +\beta )}=\;\frac{\alpha \beta t^{2}}{2(\alpha +\beta )}$

${L}=$ Length of escalator ${V}_{{b} / {esc}}=\frac{{L}}{{t}_{1}}$ When only escalator is moving. ${V}_{{esc}}=\frac{{L}}{{t}_{2}}$ when both are moving ${V}_{{b} / {g}}={V}_{{b} / {csc}}+{V}_{{csc}}$ ${V}_{{b} / {g}}=\frac{{L}}{{t}_{1}}+\frac{{L}}{{t}_{2}}$ $\Rightarrow[{t}={{L}}/{{V}_{{b} / {g}}}=\frac{{t}_{1} {t}_{2}}{{t}_{1}+{t}_{2}}]$

$\text{Av}^2$ versus x graph is shown below. Using equation of motion, $v^{2}=u^{2}+2 a x$ ...(i) We know that, the equation of straight line. y = mx + c ...(ii) Comparing Eqs. (i) and (ii), Slope,$m=2 a=\frac{80-20}{30-0}=\frac{60}{30}=2$ Now, the acceleration, $a=\frac{2}{2}=1 {\text{m}} / {\text{s}}^{2}$

Since, the particle is moving with constant acceleration. Acceleration = Constant, (the acceleration-time graph is parallel to the time axis, and acceleration is positive) $a=\;\frac{d v}{d t}$As we know, $\;\frac{d v}{d t}=C \Rightarrow ∫ d v=C ∫ d t$ $\Rightarrow \;\; v= {Ct}+C_{0}$ The velocity-time graph is straight line with positive slope $V=\;\frac{d x}{d t}$ As we know, $\;\frac{d x}{d t}=C t+C_{0}$ $\Rightarrow \;\; ∫ d x=∫(C t+C_{0}) d t \Rightarrow x=\;\frac{C t^{2}}{2}+C_{0} t+C_{1}$ This equation shows that the distance-time graph is upward concave parabola.Hence, the correct option is (b).

Given, equation of velocity, $v \;\; =v_{0}+g t+F t^{2}$ $\;\text{ At }\; t \;\; =0, x=0$ $\;\text{ At }\; t \;\; =1 s x=?$ We know that, $v=\;\frac{d x}{d t}$ $d x=v d t$ Integrating the above equation with proper limits, we get $∫_{0}^{x} d x=∫_{0}^{t-1} v d t$ $x-0=∫_{0}^{t-1}(v_{0}+g t+F t^{2}) d t$ \Rightarrow \;\; x=\[v_{0} t+;\frac{g t^{2}}{2}+;\frac{F t^{3}}{3}]{0}^{1} ⇒ x=$[v{0}(1)+;\frac{g(1)^{2}}{2}+;\frac{F(1)^{3}}{3}]$-0$$\Rightarrow ;; x=[v_{0}+;\frac{g}{2}+;\frac{F}{3}]$

${V}=\alpha {t}+\beta {t}^{2}$ $\frac{{ds}}{{dt}}=\alpha {t}+\beta {t}^{2}$ $\int _{{S}_{1}} ^{{s}_{2}} {ds}=\int _{1} ^{2}(\alpha {t}+\beta {t}^{2}) {dt}$ {S}_{2}-{S}_{1}=\[\frac{\alpha {t}^{2}}{2}+\frac{\beta {t}^{3}}{3}]_{1}^{2}$As particle is not changing direction So distance$=$displacement. Distance$=[\frac{\alpha [4-1]}{2}+\frac{\beta [8-1]}{3}]$$=\frac{3}{2} \alpha +\frac{7}{2} \beta $

From the given velocity-displacement graph, for $0 \le x \le 200$, the graph is a straight line i.e., varying linearly. As, the equation of straight line is $y=m x+c$ Here, it can be given as $v=m x+c$ where, $m=$ slope $=\;\frac{(50-10)}{200-0}=\;\frac{1}{5}$ and $c=10$ $\Rightarrow \;\; v=\;\frac{x}{5}+10$ As, acceleration, $a=\;\frac{d v}{d t}=\;\frac{v d v}{d x}$ $a=(\;\frac{x}{5}+10) \;\frac{d}{d x}(\;\frac{x}{5}+10)$ [using Eq. (i)] $=(\;\frac{x}{5}+10)(\;\frac{1}{5})=2+\;\frac{x}{25}$ For $x=0, a=2+\;\frac{0}{25}=2 {ms}^{-2}$ and for $x=200, a=2+\;\frac{200}{25}=10 {ms}^{-2}$ Therefore, the graph will have straight line till $x=200$ and for $x>200 ; v=$ constant and $a=0$ So, from the graphs given in the options, the above conditions are satisfied in option (a) only.

${t}={mx}^{2}+{n} {x}$ $\frac{1}{{v}}=\frac{{dt}}{{dx}}=2 {m} {x}+{n}$ ${v}=\frac{1}{2 {mx}+{n}}$ $\frac{{d} v}{{dt}}=-\frac{2 {m}}{(2{m} {x}+{n})^{2}}(\frac{{d} {x}}{{dt}})$ ${a}=-(2 {m}) {v}^{3}$

For the given velocity-displacement graph,Intercept, $c=v_{0}$ Slope, $m=-\;\frac{v_{0}}{x_{0}}$ $\therefore$ Slope, $m=\;\frac{d v}{d x}=-\;\frac{v_{0}}{x_{0}}$....(i) For given diagram, the equation of line, $y=m x+c$ $v=-\;\frac{v_{0}}{x_{0}} x+v_{0}$ ...(ii) Acceleration, $a=\;\frac{d v}{d x} \times \;\frac{d x}{d t}$ $a=v \;\frac{d v}{d x}$ $a=(-\;\frac{v_{0}}{x_{0}} x+v_{0})(-\;\frac{v_{0}}{x_{0}})$ [From Eqs. (i) and (ii)] $\Rightarrow \;\; a=(\;\frac{v_{0}}{x_{0}})^{2} x-\;\frac{v_{0}^{2}}{x_{0}}$ Hence, the intercept is negative and the slope is positive, so the correct graph is given in option (c).

Velocity of rubber ball when it strikes to the ground, $v_{0}=\sqrt{2 {gh}_{0}}$ Using the formula of coefficient of restitution, ${e}=\;\frac{\;\text{ velocity after collision }\;}{\;\text{ velocity before collision }\;}$ $\Rightarrow \;\; e=\;\frac{v_{1}-0}{v_{0}-0} \Rightarrow e=\;{v_{1}}/{\sqrt{2 g h_{0}}}$ $\Rightarrow \;\; v_{1}=e \sqrt{2 g h_{0}}$...(i) As, initial height of the ball $=h_{0}$. $\therefore$ The first height of the rebound, $h_{1}=\;\frac{v_{1}^{2}}{2 g}$ $\Rightarrow \;\; h_{1}=e^{2} h_{0}$ [Using Eq. (i)] The nth height of the ball to the rebound, $h_{n}=\;\frac{v_{n}^{2}}{2 g} \Rightarrow h_{n}=e^{2 n} h_{0}$ The velocity of the ball after $n$th rebound, $v_{n}=e^{n} v_{0}$ Now, the total distance travelled by the ball after nth rebound is $H=h_{0}+2 h_{1}+2 h_{2}+2 h_{3} . . .$ $H=h_{0}+2 e^{2} h_{0}+2 e^{4} h_{0}+2 e^{6} h_{0} . . .$ $H=h_{0}[1+2 e^{2}(1+e^{2}+e^{4}+e^{6} . . .)]$ Using the formula, $1+e^{2}+e^{4}+. . .=\;\frac{1}{1-e^{2}}$ $\Rightarrow \;\; H=h_{0}[1+2 e^{2}(\;\frac{1}{1-e^{2}})]$ $\Rightarrow \;\; H=h_{0}(\;\frac{1+ {e}^{2}}{1- {e}^{2}})$ $\Rightarrow \;\; H=5(\;\frac{1+(0.81)}{1-(0.81)}) \;\; ( \because e^{2}=\;\frac{h_{1}}{h_{0}}=\;\frac{81}{100})$ $H=47.6 {m}$ Now, the total time taken by the ball to come to rest $T=t_{0}+2 t_{1}+2 t_{2}+. . .$ $T=\sqrt{\;\frac{2 h_{0}}{g}}+2 \sqrt{\;\frac{2 h_{1}}{g}}+2 \sqrt{\;\frac{2 h_{2}}{g}}+. . .$ $T=\sqrt{\;\frac{2 h_{0}}{g}}[1+2 e+2 e^{2}+. . .]$ $T=\sqrt{\;\frac{2 h_{0}}{g}}[1+2 e(1+e+e^{2}+. . .)]$ $T=\sqrt{\;\frac{2 h_{0}}{g}}(\;\frac{1+e}{1-e}) \Rightarrow T=\sqrt{\;\frac{2(5)}{10}}(\;\frac{1+\sqrt{0.81}}{1-\sqrt{0.81}})$ $T=19 {s}$ The average velocity, $v_{\;\text{avg }\;}=\;\frac{H}{T}=\;\frac{47.6}{19}$ $=2.5 {m} / {s}$ Hence, the average velocity is $2.5 \frac{m}{s}$.

Let the total height of building be $x$. $T A \;\; =5 m$ $T B \;\; =25 m$ $\therefore \;\; A G=x-5$ and $B C=x-25$ For initial conditions, from second equation of motion under gravity, $s=u t+1 / 2 g t^{2}$ where, $g=10 {ms}^{-2}$ $\therefore \;\; 5=0+1 / 2 \times 10 t^{2} \Rightarrow t=1 {s}$ Now, by first equation of motion under gravity, $v_{A} \;\; =u+g t$ $\;\; =0+10=10 {ms}^{-1}$ From second equation of motion, $x-5=v_{A} t+1 / 2 g t^{2}$ . . . (i) Similarly, $x-25=1 / 2 {gt}^{2}$ Put the above value in Eq. (i), we get $x-5 \;\; =10 t+x-25$ $20 \;\; =10 t \Rightarrow t=2 {s}$ Put the value of $t$ in Eq. (i), we get $x-5 \;\; =10 \times 2+1 / 2 \times 10 \times 4$ $x-5 \;\; =20+20$ $x \;\; =45 {m}$

Velocity at ground (means zero height) is non- zero therefore one is incorrect and velocity versus height is non-linear therefore two is also incorrect. $\ {v}^{2}=2 \ {gh}$ ${v} \frac{{dv}}{ {dh}}=2 {g}= {\text{const}.}$ $\frac{{dv}}{ {dh}}= \frac{\text{ constant }}{\ {v}}$ Here we can see slope is very high when velocity is low therefore at Maximum height the slope should be very large which is in option 3 and as velocity increases slope must decrease there for option 3 is correct.

$x=\sqrt{at^{2+}2bt+c}$Differentiating w.r.t. time$\frac{dx}{dt}=v$ $=1/{2\sqrt{at^{2+}2bt+c}}\times (2at+2b)$ $\Rightarrow v={at+b}/x$ $\Rightarrow vx=at+b$Differentiating w.r.t. x$\Rightarrow \frac{dv}{dx}\times x+v=a\times \frac{dt}{dx}$Multiply both side by v$\Rightarrow (v\frac{dv}{dx})x+v^2=a$ $\Rightarrow a'x=a-v^2$ [Here a' is acceleration] $\Rightarrow a'x=a-({at+b}/x)^2$ $\Rightarrow a'x={ax^{2-}(at+b)^2}/x^2$ $\Rightarrow a'x={a(at^{2+}2bt+c)-(at+b)^2}/x^2$ $\Rightarrow a'={ac-b^2}/x^3$ $\therefore a' \propto 1/x^3$ $\therefore n=3$

${OS}=4+ \frac{1}{3}= \frac{13}{3}$ ${SD}=2- \frac{1}{3}= \frac{5}{3}$ Area of OABS is A $_{1}$ Area of ${SCD}$ is ${A}_{2}$ Distance $=|{A}_{1} |+|{A}_{2} |$ ${A}_{1}= \frac{1}{2}|\frac{13}{3}+1 ] 4=\frac{32}{3}$ ${A}_{2}= \frac{1}{2} \times \frac{5}{3} \times 2= \frac{5}{3}$ Distance $=|{A}_{1} |+|{A}_{2} |$ $= \frac{32}{3}+ \frac{5}{3}$ $= \frac{37}{3}$

Distance $=∫{v} \ {dt}$ Area under graph $= \frac{1}{2} \\times 5 \\times 8=20$

Velocity of man with respect to ground ${{V} ^{\to }}_{ {m} / {g}}={\vec{V}}_{ {m} / {A}}+{\vec{V}}_{ {A}}$ $=-1.8+36$ Velocity of man w.r.t. B ${\vec{V}}_{ {m} / {B}}={\vec{V}}_{ {m}}-{\vec{V}}_{ {B}}$ $=-1.8+36-(-72)$ $=106.2 {km} / {hr}$ $=29.5 {m} / {s}$

$v^{2}=u^{2}+2 a s$ $v^{2}=2 g h+2 g h$ $v=\\sqrt{4 g h}$ $\Rightarrow \sqrt{4 {gh}}=\\sqrt{2 {gh}}+ {gt}$ $\Rightarrow {t}=\sqrt{\frac{4 {h}}{ {g}}}-\sqrt{\frac{2 {h}}{ {g}}}$ $\Rightarrow 3.4 \sqrt{\frac{{h}}{ {g}}}$