Motion In A Straight Line Practice Questions

Given $t=a x^{2+}b x$ Differentiate with respect to time $(t)$ $\;\frac{d}{d t}(t)=a \;\frac{d}{d t} (x^2) +b \;\frac{d x}{d t}$ $=a .2 x \;\frac{d x}{d t}+b \cdot \;\frac{d x}{d t}$ $1=2 a x v+b v=v(2 a x+b)$[as $\;\frac{d x}{d t}=v]$ $\Rightarrow 2 a x+b=\;\frac{1}{v}$ Again differentiating,$\;2 a \;\frac{d x}{d t}+0=-\;\frac{1}{v^2} \;\frac{d v}{d t}$ $\;\Rightarrow \;\frac{d v}{d t}=f=-2 a v^3$ $\;\;\text{ (as }\; \;\frac{d v}{d t}=f=\;\text{ Acceleration) }\;$

Initially car starts from rest so ${\text{u}}=0$.Now distance from $A$ to $B$,$S\;=\;\frac{1}{2} f t_1^2$ $\Rightarrow f t_1^2\;=2 S$Distance from $B$ to $C= (f t_1) t$In ${\text{B}}$ to ${\text{C}}$ velocity is constant and ${\text{v}}=f t_1$Distance from $C$ to $D$ $\; \;\; =\;\frac{u^2}{2 a}=\;\frac{ (f t_1) ^2}{2(f / 2)}=f t_1^2=2 S$ $\;\Rightarrow S+f t_1 t+2 S=15 S$ $\;\Rightarrow f t_1 t=12 S . . . . . .(1)$ $\;\;\text{ But }\; \;\; \;\frac{1}{2} f t_1^2=S . . . . . .(2)$ $\;\;\text{ On dividing the above two equations, we get }\; t_1=\;\frac{t}{6}$ $\;\Rightarrow S=\;\frac{1}{2} f (\;\frac{t}{6}) ^2=\;\frac{f t^2}{72}$

Assume $a$ be the retardation for both the vehicle thenIn case of automobile,$\;u_1^{2-}2 a s_1=0$ $\;\Rightarrow u_1^2=2 a s_1$And in case for car,$\;u_2^2=2 a s_2$ $\; \therefore (\;\frac{u_2}{u_1}) ^2=\;\frac{s_2}{s_1}$ $\;\Rightarrow (\;\frac{120}{60}) ^2=\;\frac{s_2}{20}$ $\;\Rightarrow s_2=80 {\text{m}}$

We know that equation of motion, $s=u t+\;\frac{1}{2} g t^2$, Initial speed of ball is zero and it take T second to reach the ground. $\therefore h=\;\frac{1}{2} g T^2$ After $T {/} 3$ second, vertical distance moved by the ball $h^{'}=\;\frac{1}{2} g (\;\frac{T}{3}) ^2$ $\Rightarrow h^{'}=\;\frac{1}{2} \times \;\frac{8 T^2}{9}$ $=\;\frac{h}{9}$ $\therefore$ Height from ground$=h-\;\frac{h}{9}=\;\frac{8 h}{9}$

For case 1 :${\text{u}}=50 \frac{\text{km}}{\text{hr}}=\;\frac{50 \times 1000}{3600} \frac{\text{m}}{\ s}=\;\frac{125}{9} \frac{\text{m}}{\ s}, {\text{v}}=0, {\ s}=6 {\text{m}}, a=\;\text{ ? }\;$ $\therefore 0^2=u^{2+}2 a s$ $\Rightarrow a=-\;\frac{u^2}{2 s}$ $\Rightarrow a=-\;\frac{ (\frac{126}{9}) ^2}{2 \times 6}=-16 \frac{\text{m}}{\ s}^2$ For case 2 :${\text{u}}=100 \frac{\text{km}}{\text{hr}}=\;\frac{100 \times 1000}{3600} \frac{\text{m}}{\ s}=\;\frac{250}{9} \frac{\text{m}}{\ s}, {\text{v}}=0, a=-16, {\ s}=?$ $\therefore 0^2=u^{2+}2 a s$ $\Rightarrow s=-\;\frac{u^2}{2 a}$ $\Rightarrow s=-\;\frac{ (\frac{250}{9}) ^2}{2 \times -16}=24 {\text{m}}$

Given that $x=\alpha t^3 \;\;$ and $y=\beta t^3$ $\therefore v_x=\;\frac{d x}{d t}=3 \alpha t^2$and $\;\; v_y=\;\frac{d y}{d t}=3 \beta t^2$ $\; \therefore v={\sqrt{v_x^{2+}v_y^2}}$ $\;={\sqrt{9 \alpha ^2 t^{4+}9 \beta ^2 t^4}}$ $\;=3 t^2 {\sqrt{\alpha ^{2+}\beta ^2}}$

Assume the initial velocity of each particle is $=u$ And height of building $= {\text{h}}$ If final velocity of $A$ is $v_A$ then $v_A{ }^2=u^{2+}2(-g)(-h)=u^{2+}2 g h$ If final velocity of $B$ is $v_B$ then $v_B^2=u^{2+}2 g$ h $\therefore {\text{V}}_{ {\text{A}}}= {\text{V}}_{ {\text{B}}}$ Sign Rule : Take the direction of initial velocity positive opposite direction as negative. Here for ball A initial velocity u is upward so upward is positive and downward is negative. That is why gravity is $=-g$ and height $=-h$ And for ball $B$ initial velocity $u$ is downward so downward is positive and upward is negative. That is why gravity is $=+g$ and height $=+h$

Given that initial speed of two cars $u$ and $4 u$ and final speed $v$ is 0 for both car. Both car is gradually slowing down so acceleration $=(-a)$ So formula becomes, $0=u^{2-}2 a$ s. $\Rightarrow u^2=2 a s$ For first car $u^2=2 a s_1$.....(i) For second car $(4 u)^2=2 a s_2$ ....(ii) Dividing $(i)$ and $(i i)$, $\;\frac{u^2}{16 u^2}=\;\frac{2 a s_1}{2 a s_2}$ $\Rightarrow \;\frac{1}{16}=\;\frac{s_1}{s_2}$