Relative Motion in One Dimension Questions in English

(D) Given velocities are $v_1 = 6 \, m/s$ and $v_2 = 10 \, m/s$.
Since both bodies move in the same direction from the same point,their relative velocity is $v_{rel} = v_2 - v_1 = 10 - 6 = 4 \, m/s$.
The separation between them at any time $t$ is given by $s = v_{rel} \times t$.
We are given the separation $s = 40 \, m$.
Substituting the values: $40 = 4 \times t$.
Therefore,$t = \frac{40}{4} = 10 \, s$.

(D) The speed of the train with respect to the ground is $V_T = 90 \,km/h$.
Converting this to $m/s$: $V_T = 90 \times \frac{5}{18} = 25 \,m/s$.
The speed of the person with respect to the train is $V_{P/T} = 1 \,m/s$ in the direction of the train's motion.
The speed of the person with respect to the ground $(V_P)$ is given by the relative velocity formula: $V_P = V_{P/T} + V_T$.
Substituting the values: $V_P = 1 \,m/s + 25 \,m/s = 26 \,m/s$.
Therefore,the speed of the person with respect to the ground is $26 \,m/s$.

(C) Let the man run with a constant speed $v$. Let $t$ be the time taken to catch the bus.
The distance covered by the bus in time $t$ is $s_b = \frac{1}{2} a t^2 = \frac{1}{2} \times 3 \times t^2 = 1.5 t^2$.
The total distance the man must cover to reach the bus is $s_m = 6 + s_b = 6 + 1.5 t^2$.
Since the man runs with constant speed $v$,the distance he covers is $s_m = v t$.
Equating the two expressions for $s_m$: $v t = 6 + 1.5 t^2$,which rearranges to $1.5 t^2 - v t + 6 = 0$.
For the man to catch the bus,the time $t$ must be a real value. This requires the discriminant of the quadratic equation to be greater than or equal to zero $(D \ge 0)$.
$D = (-v)^2 - 4(1.5)(6) \ge 0$
$v^2 - 36 \ge 0$
$v^2 \ge 36 \implies v \ge 6\,m s^{-1}$.
Thus,the minimum speed required is $6\,m s^{-1}$.

(C) Let the two balls meet after time $t$ seconds.
For the ball dropped from the top (downward motion):
The distance covered is $s_1 = \frac{1}{2} g t^2$.
For the ball thrown upwards from the bottom:
The distance covered is $s_2 = u t - \frac{1}{2} g t^2$,where $u = 40\,m/s$.
Since the total height of the building is $100\,m$,the sum of the distances covered by both balls must equal the height of the building:
$s_1 + s_2 = 100$
Substituting the expressions:
$\frac{1}{2} g t^2 + (40 t - \frac{1}{2} g t^2) = 100$
$40 t = 100$
$t = \frac{100}{40} = 2.5\,s$.
Thus,the balls will meet after $2.5\,s$.

(A) First,convert the speeds from $km/h$ to $m/s$ by multiplying by $\frac{5}{18}$.
$V_A = 72 \times \frac{5}{18} = 20 \ m/s$ (towards North,taken as positive).
$V_B = 108 \times \frac{5}{18} = 30 \ m/s$ (towards South,taken as negative,so $V_B = -30 \ m/s$).
Velocity of train $B$ with respect to $A$ is $V_{BA} = V_B - V_A = -30 - 20 = -50 \ m/s$.
Velocity of ground with respect to $B$ is $V_{GB} = V_G - V_B = 0 - (-30) = 30 \ m/s$.
Thus,the velocities are $-50 \ m/s$ and $30 \ m/s$.

(C) Let the two cars be $A$ and $B$. The initial velocity of car $A$ is $u_A = 20 \,m \,s^{-1}$ and car $B$ is $u_B = -20 \,m \,s^{-1}$.
The relative initial velocity of car $B$ with respect to car $A$ is $u_{BA} = u_B - u_A = -20 - 20 = -40 \,m \,s^{-1}$.
The magnitude of relative initial velocity is $|u_{BA}| = 40 \,m \,s^{-1}$.
Both cars retard at $a = 2 \,m \,s^{-2}$. Let $a_A = -2 \,m \,s^{-2}$ and $a_B = 2 \,m \,s^{-2}$.
The relative acceleration of car $B$ with respect to car $A$ is $a_{BA} = a_B - a_A = 2 - (-2) = 4 \,m \,s^{-2}$.
Using the equation of motion $v^2 = u^2 + 2as$ for relative motion, where final relative velocity $v_{BA} = 0$:
$0^2 = (-40)^2 + 2(-4)S$
$0 = 1600 - 8S$
$8S = 1600 \implies S = 200 \,m$.
This is the relative distance covered by the cars before coming to rest.
The initial distance between the cars was $300 \,m$.
Therefore, the remaining distance between them when they come to rest is $300 \,m - 200 \,m = 100 \,m$.

(C) To avoid a collision,the relative velocity of car $A$ with respect to car $B$ must become zero before or at the moment the distance between them becomes zero.
Let $u_r$ be the initial relative velocity: $u_r = V_A - V_B$.
Let $a_r$ be the relative acceleration: $a_r = -a - 0 = -a$.
Using the kinematic equation $v^2 = u^2 + 2as$,where $v$ is the final relative velocity $(0)$,$u$ is the initial relative velocity $(V_A - V_B)$,and $s$ is the relative displacement $S'$ required to stop:
$0^2 = (V_A - V_B)^2 + 2(-a)S'$.
Solving for $S'$,we get $S' = \frac{(V_A - V_B)^2}{2a}$.
For no collision to occur,the initial distance $S$ must be greater than or equal to the stopping distance $S'$.
Therefore,$S \geqslant \frac{(V_A - V_B)^2}{2a}$.

(C) Let $h$ be the height of the escalator.
Speed of the girl walking on the stationary escalator is $v_{g} = \frac{h}{20}$.
Speed of the moving escalator is $v_{e} = \frac{h}{30}$.
When the girl walks on the moving escalator, her effective speed is the sum of her walking speed and the escalator's speed: $v_{total} = v_{g} + v_{e}$.
$v_{total} = \frac{h}{20} + \frac{h}{30} = h \left( \frac{3+2}{60} \right) = \frac{5h}{60} = \frac{h}{12}$.
The time taken to cover the distance $h$ is $t = \frac{h}{v_{total}} = \frac{h}{h/12} = 12 \,s$.

(C) The given situation is as shown in the figure.
Velocity of bus $A$,$v_A = 36 \ km/h = 36 \times \frac{5}{18} = 10 \ m/s$ (towards east).
Velocity of bus $B$,$v_B = 18 \ km/h = 18 \times \frac{5}{18} = 5 \ m/s$ (towards west).
Let the direction towards the east be positive. Then,$v_A = +10 \ m/s$ and $v_B = -5 \ m/s$.
The relative velocity of bus $B$ with respect to bus $A$ is given by:
$v_{BA} = v_B - v_A = (-5) - (10) = -15 \ m/s$.
The negative sign indicates that the bus $B$ appears to move towards the west relative to bus $A$.
Therefore,bus $B$ appears to bus $A$ as moving with a speed of $15 \ m/s$ from east to west.

(A) The motion of two cars $A$ and $B$ is shown in the figure.
Relative velocity of car $A$ with respect to car $B$ is:
$v_{AB} = v_A - v_B = 15 \,m/s - 10 \,m/s = 5 \,m/s$
To completely overtake car $B$, car $A$ must cover a total distance equal to the sum of the lengths of both cars:
$s = 4 \,m + 4 \,m = 8 \,m$
The time taken to complete the overtake is given by:
$t = \frac{s}{v_{AB}} = \frac{8 \,m}{5 \,m/s} = 1.6 \,s$

(C) Let the police party catch the thief after time $t$.
Distance traveled by the police party in time $t$ is $d_p = v t$.
Distance traveled by the thief in time $t$ starting from rest with acceleration $a$ is $d_t = x + \frac{1}{2} a t^2$.
For the police to catch the thief,the distance traveled by the police must be greater than or equal to the distance traveled by the thief:
$v t \geq x + \frac{1}{2} a t^2$
Rearranging the terms,we get a quadratic inequality in $t$:
$\frac{1}{2} a t^2 - v t + x \leq 0$
For a real solution for $t$ to exist,the discriminant of the corresponding quadratic equation $\frac{1}{2} a t^2 - v t + x = 0$ must be greater than or equal to zero.
The discriminant $D = b^2 - 4ac$ is given by:
$D = (-v)^2 - 4(\frac{1}{2} a)(x) \geq 0$
$v^2 - 2 a x \geq 0$
$v^2 \geq 2 a x$