Kinematics-1D

In OPTIONS (A), (C) and (D), the given graphs represent uniformly decelerated motion of a particle in a straight line with positive initial velocity. Distance-time graph of such a motion is shown here.

Hence, the correct answer is (B).

Distance travelled by car in 15 s is
$s _ {1} = \frac {1}{2} \times A C \times O C = \frac {1}{2} (4 5) (1 5) = \frac {6 7 5}{2} \mathrm{m}$
Distance travelled by scooter in 15 s is
$s _ {2} = v \times t = 3 0 \times 1 5 = 4 5 0 \mathrm{m}$
Required difference in distance is $(s_{2}-s_{1})=\Delta s$
$\Rightarrow \quad \Delta s = 4 5 0 - \frac {6 7 5}{2} = \frac {2 2 5}{2} = 1 1 2. 5 \mathrm{cm}$
Let the car catch scooter in time t, then
$\begin{array}{r l} & {\frac {6 7 5}{2} + 4 5 (t - 1 5) = 3 0 t} \\ {\Rightarrow} & {3 3 7. 5 + 4 5 t - 6 7 5 = 3 0 t} \\ {\Rightarrow} & {1 5 t = 3 3 7. 5} \\ {\Rightarrow} & {t = 2 2. 5 \mathrm{s}} \end{array}$
Hence, the correct answer is (D).

Using, $v^{2} = u^{2} - 2as$ , we get
$\begin{array}{r l} & 0 = u ^ {2} - 2 a s \\ \Rightarrow & s = \frac {u ^ {2}}{2 a} \\ \Rightarrow & \frac {s _ {1}}{s _ {2}} = \frac {u _ {1} ^ {2}}{u _ {2} ^ {2}} \\ \Rightarrow & s _ {2} = \left(\frac {u _ {2}}{u _ {1}}\right) ^ {2} s _ {1} \\ \Rightarrow & s _ {2} = (2) ^ {2} (4 0) = 1 6 0 \mathrm{m} \\ & \text { Hence, the correct answer is (C). } \end{array}$

Time taken by car to reach at location $P$ from location $Q$ is
$\begin{array}{r l} & t = \frac {Q R}{v} + \frac {R P}{(v / 2)} \\ \Rightarrow & t = \frac {(L - x)}{v} + \frac {2 \sqrt {d ^ {2} + x ^ {2}}}{v} \end{array}$
For t to be minimum, we have

$\begin{array}{l l} & \frac {d t}{d x} = 0 \\ \Rightarrow & \frac {d t}{d x} = \frac {1}{v} (0 - 1) + 2 \times \left(\frac {1}{2}\right) \frac {1}{v} \times \frac {2 x}{\sqrt {d ^ {2} + x ^ {2}}} \\ \Rightarrow & \frac {d t}{d x} = \frac {- 1}{v} + \frac {2 x}{v \sqrt {d ^ {2} + x ^ {2}}} \end{array}$
Since, for minimum value of t
$\begin{array}{r l} & {\frac {d t}{d x} = 0} \\ {\Rightarrow} & {- \frac {1}{v} + \frac {2 x}{v \sqrt {d ^ {2} + x ^ {2}}} = 0} \\ {\Rightarrow} & {1 = \frac {2 x}{\sqrt {d ^ {2} + x ^ {2}}}} \\ {\Rightarrow} & {4 x ^ {2} = d ^ {2} + x ^ {2}} \\ {\Rightarrow} & {3 x ^ {2} = d ^ {2}} \\ {\Rightarrow} & {x = \frac {d}{\sqrt {3}}} \end{array}$
Hence, the correct answer is (B).

As in distance vs time graph slope is equal to speed in the given graph slope increase initially which is incorrect.

Ans. (C)
Density of sphere is $\rho = \frac{m}{v} = \frac{3m}{4\pi R^3}$
$\begin{array}{l} \text {Since,} \frac {1}{\rho} \frac {d \rho}{d t} = \text {constant} \\ \Rightarrow \quad \frac {4 \pi R ^ {3}}{3 m} \times \frac {3 m}{4 \pi} \left(\frac {- 3}{R ^ {4}}\right) \frac {d R}{d t} = \text {constant (Let K)} \\ \Rightarrow \quad \frac {- 3}{R} \frac {d R}{d t} = K \quad \Rightarrow \quad \frac {d R}{d t} = \left(\frac {- K}{3}\right) R \\ \therefore \quad \frac {d R}{d t} \propto R \end{array}$
Velocity v of any point on the surface of sphere is equal to $\frac{dR}{dt}$ (rate of change of radius of outer layer).

Velocity of the body going upwards is given by $v = v_{0} - gt$ , where $v_{0}$ is the initial velocity.
Hence, the graph between velocity and time should be a straight line with negative slope $(g)$ and intercept $v_{0}$ .
Also, during the whole motion, acceleration of the body is constant i.e., slope should be constant.
Hence, the correct answer is (C).

Here, acceleration is given by, a = -c
$\Rightarrow \frac {d v}{d t} = - c$
$\Rightarrow \frac {d x}{d t} \cdot \frac {d v}{d x} = - c$
$\Rightarrow v \frac {d v}{d x} = - c$
$\Rightarrow v d v = - c d x$
$\Rightarrow \frac {v ^ {2}}{2} = - c x + k$
$\Rightarrow x = - \frac {v ^ {2}}{2 c} + \frac {k}{c}$
Hence, the correct answer is (A).

C

Acceleration of car, $a_{C} = 4 \, ms^{-2}$
Acceleration of bus, $a_{B} = 2 \, ms^{-2}$
Initial separation between the bus and car is
$s _ {C B} = 2 0 0 \mathrm{m}$
Acceleration of car with respect to bus is
$a _ {C B} = a _ {C} - a _ {B} = 2 \mathrm{ms} ^ {- 2}$
Initial velocity of car w.r.t. bus is
$u _ {C B} = 0, t = ?$
$\begin{array}{r l} & {\mathrm{Since,} s _ {C B} = u _ {C B} t + \frac {1}{2} a _ {C B} t ^ {2}} \\ & {\Rightarrow 2 0 0 = (0) t + \frac {1}{2} (2) t ^ {2}} \\ & {\Rightarrow t ^ {2} = 2 0 0} \\ & {\Rightarrow t = 1 0 \sqrt {2} \mathrm{s}} \end{array}$
Hence, the correct answer is (C).

$h$ is height to reach from one floor to other floor speed of walking $= \frac{h}{t_1} = V_1$ speed of escalator $= \frac{h}{t_2} = V_2$ so time taken when escalator is walking and she is also walking on it, $t = \frac{h}{V_1 + V_2}$ $\frac{1}{t} = \frac{V_1 + V_2}{h} = \frac{1}{t_1} + \frac{1}{t_2} \Rightarrow t = \frac{t_1 t_2}{t_1 + t_2}$ .

Ans. (1)
For particle 1
$- 2 4 0 = + 1 0 t - \frac {1}{2} g t ^ {2}$
$5 t ^ {2} - 1 0 t - 2 4 0 = 0$
$t _ {1} = 8 \sec$
For particle 2
$- 2 4 0 = 4 0 t - \frac {1}{2} g t ^ {2}$
$5 t ^ {2} - 4 0 t - 2 4 0 = 0$

$t _ {2} = 1 2 \sec$
For $0 < t < 8 \sec \rightarrow a_{rel} = 0$
Straight line $x - t$ graph
For $8 < t < 12\mathrm{sec}\rightarrow a_{rel} = -g$
Downward parabola

For $t > 12\mathrm{sec}\rightarrow$ Both particles comes to rest.

Using $h = ut + \frac{1}{2} at^{2}$
For stone 1, we have
$y _ {1} = 1 0 t - \frac {1}{2} g t ^ {2}$
and for stone 2, we have
$y _ {2} = 4 0 t - \frac {1}{2} g t ^ {2}$
Relative position of the second stone with respect to the first,
$\Delta y = y _ {2} - y _ {1} = 4 0 t - \frac {1}{2} g t ^ {2} - 1 0 t + \frac {1}{2} g t ^ {2}$
$\Rightarrow \Delta y = 3 0 t$
After 8 seconds, stone 1 reaches ground, i.e., $y_{1} = -240$ m
$\Rightarrow \quad \Delta y = y _ {2} - y _ {1} = 4 0 t - \frac {1}{2} g t ^ {2} + 2 4 0$
Therefore, it will be a parabolic curve till other stone reaches ground.
Hence, the correct answer is (A).

$\overrightarrow{V_{BA}} = \overrightarrow{V_{B}} - \overrightarrow{V_{A}}$
$= (2 0 \times \cos 6 0 ^ {\circ} \hat {i} + 2 0 \sin 6 0 ^ {\circ} \hat {j}) - 1 0 \hat {i} = 1 0 \sqrt {3} \hat {j}$

$V_{A}^{2}=u^{2}-2g(-h)=u^{2}+2gh$ $V_{B}^{2}=u^{2}+2gh=u^{2}+2gh$ $V_{A}=V_{B}$ (Attain the same final velocity).

Time taken by the particle to reach the top most point is,
$t = \frac {u}{g}\tag{... (1}$

Time taken by the particle to reach the ground is
$t ^ {\prime} = n t$
$s = u t + \frac {1}{2} a t ^ {2}$
$\Rightarrow - H = u (n t) - \frac {1}{2} g (n t) ^ {2}$
Using (1), we get
$\begin{array}{r l} & {- H = u \times n \left(\frac {u}{g}\right) - \frac {1}{2} g n ^ {2} \left(\frac {u}{g}\right) ^ {2}} \\ {\Rightarrow} & {- 2 g H = 2 n u ^ {2} - n ^ {2} u ^ {2}} \\ {\Rightarrow} & {2 g H = n u ^ {2} (n - 2)} \end{array}$
Hence, the correct answer is (D).

Time to reach the maximum height $t_{1}=u/g$ If $t_{2}$ be the time taken to hit the ground $-H=ut_{2}-\frac{1}{2}gt_{2}^{2}$ But $t_{2}=nt_{1}$ (given) $\Rightarrow-H=u\frac{nu}{g}-\frac{1}{2}g\frac{n^{2}u^{2}}{g^{2}}$ $\Rightarrow 2g H = nu^{2}(n - 2).$