Motion In A Straight Line Practice Questions

a=constantV = at$S= \frac{1}{2} a t^{2}$

$x=a t+b t^{2}-c t^{3}$ $V=a+2 b t-3 c t^{2}$Accn= 2b-6ct=0$\Rightarrow t= \frac{b}{3 C}$ $V=a+ \frac{2 b^{2}}{3 C}-\ \frac{b^{2}}{3 C}=a+ \frac{b^{2}}{3 C}$

$V=b \\sqrt{x} \ \frac{d x}{d t}=b \\sqrt{x} \int _{0}^{x} \ \frac{d x}{\\sqrt{x}}$ $=b \int _{0}^{\tau } d t\; 2 \ \sqrt{x}=b \tau \\sqrt{x}=b \tau / {2 V}= \frac{b^{2} \tau }{2}$

$\vec{S}$ = $(5\hat{i}+4\hat{j})$ 2 + $1/2 (4\hat{i}+4\hat{j})$ 4 = $10\hat{i}+8\hat{j}+8\hat{i}+8\hat{j}$ r_\vec{f}-r_\vec{i} = $18\hat{i}+16\hat{j}$ r_\vec{f} = $20\hat{i}+20\hat{j}$ |r_\vec{f}| = 20 $\sqrt2$

To completely cross the freight train, the passenger train must cover a total distance $D = L_p + L_f = 60 + 120 = 180 \text{ m}$.
When moving in the same direction, the relative speed is $V_1 = 80 - 30 = 50 \text{ km/hr}$.
When moving in opposite directions, the relative speed is $V_2 = 80 + 30 = 110 \text{ km/hr}$.
Using the formula $t = \frac{D}{V}$, the ratio of times is $\frac{t_1}{t_2} = \frac{D/V_1}{D/V_2} = \frac{V_2}{V_1}$.
Substituting the values, we get $\frac{t_1}{t_2} = \frac{110}{50} = \frac{11}{5}$.

\ \vec{V}=30 t \ \hat{i}-40 t \ {j} ^{\^} \ \vec{a}=30 \ {i}^{\^ }-40 \ {j} ^{\^} $|a|=\\sqrt{30^{2}+40^{2}}=50$

Initial and final slopes are zero (so option '2' is wrong) all other are correct

At $t = 15$ s, the distance traveled by the car is the area of the velocity-time triangle: $S_c = \frac{1}{2} \times 15 \times 45 = 337.5$ m, while the scooter's distance is $S_s = 30 \times 15 = 450$ m.
The difference between these distances is $S_s - S_c = 450 - 337.5 = 112.5$ m.
For $t > 15$ s, the car catches up when their distances are equal, expressed as $S_c(15) + v_{\text{car}}(t - 15) = v_{\text{scooter}}t$.
Substituting the values, we get $337.5 + 45(t - 15) = 30t$, which simplifies to $337.5 + 45t - 675 = 30t$.
Rearranging for $t$ gives $15t = 337.5$, leading to $t = \frac{337.5}{15} = 22.5$ s.

$\;\text{ Since stopping distance, }\; s =\;\frac{ v ^{2}}{2 a }$ $\;\text{ For case }\; 1,40 =\;\frac{(40 \times \frac{5}{18})^{2}}{2 a } . . . . . . . . (1)$ $\;\text{ For case }\; 2, s=\;\frac{(80 \times \frac{5}{18})^{2}}{2 a} . . . . . . . (2)$ $\;\text{ From }\;(1) \;\text{ and }\;(2),$ $s=160 m$

Acceleration of Car, ${a}_{ {C}}=4 {m} / {s}^{2}$ Acceleration of bus, ${a}_{ {B}}=2 {m} / {s}^{2}$ Initial distance between them, ${S}=200 {m}$ Acceleration of Car with respect to bus, ${a}_{ {CB}}= {a}_{ {C}}- {a}_{ {B}}=4-2=2 {m}/ {s}^{2}$ As initially both are at rest ${so}_{0}, {u}_{ {CB}}=0$ $\therefore {S}= {U}_{ {CB}} \times {t}+\;\frac{1}{2} {a}_{ {CB}} \times {t}^{2}$ $\Rightarrow 200=0+\;\frac{1}{2} \times 2 \times t^{2}$ $\Rightarrow t^{2}=200$ $\Rightarrow {t}=10 \sqrt{2} {sec} .$

Velocity at any time is given byv = u +at$v=v_0+(-g)t$ $v=v_0-gt$⇒straight line with negative slope

$a=-C$ $\frac{VdV}{dx}=-C$ $VdV=-CdX$ $V^2/2=-Cx+k$ $x=-V^2/{2C}+K/C$

$y_1=10t-1/2gt^2$ $y_2=40t-1/2gt^2$ $y_2-y_1=30t$(straight line)but stone with 10 m/s speed will fall first and the other stone is still in air.Therefore path will becomeparabolic till other stone reaches ground.

Time to reach the maximum height$t_1=u/g$If $t_2$ be the time taken to hit the ground$-H=ut_2-1/2gt_2^2$But $t_2=nt_1$ (given)$\Rightarrow -H=u{\nu}/g-1/2g{n^2u^2}/g^2$ $\Rightarrow 2gH=\nu^2(n-2)$

Given $\;\frac{d v}{d t}=-2.5 {\sqrt{v}}$ $\Rightarrow \;{d v}/{{\sqrt{v}}}=-2.5 \text{dt}$On integrating, $∫_{{6.25}^0 v}^{-1 / 2} d v=-2.5 ∫_{0}^{t} \text{dt}$ \;\Rightarrow \[;\frac{v^{1 / 2}}{ ({ }^{1 / 2}) }]$ _{6.25}^0=-2.5[t]_0^t$$;\Rightarrow -2(6.25)^{1 / 2}=-2.5 t$$;\Rightarrow t=2 {\ sec}$

For downward motion : $v=-g t$The velocity of the rubber ball increases in downward direction and we get a straight line between $v$ and $t$ with a negative slope. Also applying $y-y_0=u t+\;\frac{1}{2} a t^2$We get $y-h=-\;\frac{1}{2} g t^2 \Rightarrow y=h-\;\frac{1}{2} g t^2$ The graph between $y$ and $t$ is a parabola with $y=h$ at $t=0$. As time increases $y$ decreases. For upward motion :The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here $v=u-g t$ where $u$ is the velocity just after collision. As $t$ increases, $v$ decreases. We get a straight line between $v$ and $t$ with negative slope. Also $y=u t-\;\frac{1}{2} g t^2$All these characteristics are represented by graph $(B)$.

For the body starting from rest$x_1=0+\;\frac{1}{2} a t^2 \Rightarrow x_1=\;\frac{1}{2} a t^2$For the body moving with constant speed $\;x_2=v t$ $\; \therefore x_1-x_2=\;\frac{1}{2} a t^{2-}v t$ $\; \;\; \Rightarrow \;\frac{d (x_1-x_2) }{d t}=a t-v$at $t=0, \;\; x_1-x_2=0$, so graph should start from origin. For $a t< v$; the slope is negative that means $x_1-x_2< 0$ so initially velocity of 1 st body is less than second body and velocity of 1 st body is increasing gradually. For $a t=v ;$ the slope is zero. So $x_1-x_2=0$ it means here velocity of both the bodies are same.For $a t>v ;$ the slope is positive. So $x_1-x_2>0$ it means here velocity of first body is greater than second body.We know the relation between distance and time is.$S=u t+\;\frac{1}{2} a t^2$, which is a equation parabola. So the graph should be a parabola.These characteristics are represented by graph $(b)$.

Given that, ${\text{v}}= {\text{v}}_0+ {\text{gt}}+ {\text{ft}}^2$ We know that, $v=\;\frac{d x}{d t}$ $\Rightarrow \text{dx}=v \text{dt}$ Integrating, $∫_{0}^{x} \text{dx}=∫_{0}^{t} v \text{dt}$ or $\;\; x=∫_{0}^{t} (v_0+g t+f t^2) \text{dt}$ = \[v_0 t+;\frac{g t^2}{2}+;\frac{f t^3}{3}]$ _0^t$or$x=v_0 t+;\frac{g t^2}{2}+;\frac{f t^3}{3}$At$t=1, ;; x=v_0+;\frac{g}{2}+;\frac{f}{3}$.