Relative Velocity (river boat, rain, wind) Questions in English

(B) Let the two cars be $A$ and $B$. Since they are moving in the same direction with the same speed $v = 30 \, km/hr$,their relative velocity is $v_{BA} = v_B - v_A = 30 - 30 = 0 \, km/hr$.
Thus,the distance between them remains constant at $d = 5 \, km$.
Let the speed of the third car $C$ moving in the opposite direction be $v_C \, km/hr$.
The relative velocity of car $C$ with respect to cars $A$ and $B$ is $v_{rel} = v_C - (-30) = (v_C + 30) \, km/hr$.
The time taken to cross the two cars is given as $t = 4 \, minutes = \frac{4}{60} \, hours = \frac{1}{15} \, hours$.
Using the formula $t = \frac{d}{v_{rel}}$,we have:
$\frac{1}{15} = \frac{5}{v_C + 30}$
$v_C + 30 = 5 \times 15$
$v_C + 30 = 75$
$v_C = 75 - 30 = 45 \, km/hr$.

(B) Let $\vec{v}_{rg}$ be the velocity of rain with respect to the ground,$\vec{v}_{mg}$ be the velocity of the man with respect to the ground,and $\vec{v}_{rm}$ be the velocity of the rain with respect to the man.
When the man is at rest,the rain falls at an angle of $30^\circ$ with the vertical. This indicates the direction of $\vec{v}_{rg}$.
When the man runs at $10 \ km/hr$,the rain appears to fall vertically,meaning the horizontal component of $\vec{v}_{rm}$ is zero.
Using the relative velocity equation: $\vec{v}_{rg} = \vec{v}_{rm} + \vec{v}_{mg}$.
Equating the horizontal components: $v_{rg} \sin 30^\circ = v_{mg}$.
Given $v_{mg} = 10 \ km/hr$,we have $v_{rg} \sin 30^\circ = 10$.
Since $\sin 30^\circ = 0.5$,we get $v_{rg} = 10 / 0.5 = 20 \ km/hr$.

(C) Let $\vec{v}_{rg}$ be the velocity of rain with respect to the ground,$\vec{v}_{mg}$ be the velocity of the man with respect to the ground,and $\vec{v}_{rm}$ be the velocity of the rain with respect to the man.
From the relative velocity equation: $\vec{v}_{rg} = \vec{v}_{rm} + \vec{v}_{mg}$,which implies $\vec{v}_{rm} = \vec{v}_{rg} - \vec{v}_{mg}$.
When the man is at rest,the rain falls at $30^\circ$ to the vertical. Let $v_r$ be the magnitude of the rain's velocity. Then $\vec{v}_{rg} = v_r \sin 30^\circ \hat{i} - v_r \cos 30^\circ \hat{j}$.
When the man runs at $10 \, km/h$,his velocity is $\vec{v}_{mg} = 10 \hat{i}$.
The velocity of rain relative to the man is $\vec{v}_{rm} = (v_r \sin 30^\circ - 10) \hat{i} - v_r \cos 30^\circ \hat{j}$.
Since the rain hits the man vertically,the horizontal component of $\vec{v}_{rm}$ must be zero:
$v_r \sin 30^\circ - 10 = 0 \implies v_r (1/2) = 10 \implies v_r = 20 \, km/h$.
The speed of raindrops with respect to the man is the vertical component: $v_{rm} = v_r \cos 30^\circ = 20 \times (\sqrt{3}/2) = 10\sqrt{3} \, km/h$.

(C) Let the velocity of the boat with respect to the ground be $\vec{v}_b = 3i + 4j$.
Let the velocity of the water with respect to the ground be $\vec{v}_w = -3i - 4j$.
The relative velocity of the boat with respect to the water is given by $\vec{v}_{bw} = \vec{v}_b - \vec{v}_w$.
Substituting the given values: $\vec{v}_{bw} = (3i + 4j) - (-3i - 4j)$.
$\vec{v}_{bw} = 3i + 4j + 3i + 4j = 6i + 8j$.

(D) The length of the train is $L = 150\, m$.
The velocity of the train is $v_t = 10\, m/s$ (towards North).
The velocity of the parrot is $v_p = -5\, m/s$ (towards South,taking North as positive).
The relative velocity of the parrot with respect to the train is $v_{rel} = v_p - v_t = -5 - 10 = -15\, m/s$.
The magnitude of the relative velocity is $|v_{rel}| = 15\, m/s$.
The time taken by the parrot to cross the train is $t = \frac{L}{|v_{rel}|} = \frac{150}{15} = 10\, s$.

(A) To cross the river in the shortest time,the component of the swimmer's velocity perpendicular to the river flow must be maximized.
Let the width of the river be $d$ and the swimmer's velocity in still water be $v_s = 10\, m/min$.
The time taken to cross the river is given by $t = \frac{d}{v_s \cos \theta}$,where $\theta$ is the angle the swimmer makes with the direction perpendicular to the river flow.
To minimize $t$,$\cos \theta$ must be maximum,which occurs when $\theta = 0^\circ$.
Therefore,the swimmer should swim perpendicular to the river flow,i.e.,due north.

(C) Let $v_m$ be the velocity of the man and $v_r$ be the velocity of the river flow.
The man swims at an angle of $120^\circ$ with the direction of the river flow to reach the exactly opposite point.
This means the component of the man's velocity along the river flow must cancel out the river's velocity.
The angle between the man's velocity vector and the line perpendicular to the bank is $120^\circ - 90^\circ = 30^\circ$.
Thus,the horizontal component of the man's velocity is $v_m \sin(30^\circ)$.
For the man to reach the opposite point,this horizontal component must be equal to the river's velocity:
$v_r = v_m \sin(30^\circ)$
Given $v_m = 0.5\, m/s$ and $\sin(30^\circ) = 0.5$,
$v_r = 0.5 \times 0.5 = 0.25\, m/s$.

(C) The velocity of object $A$ is $v_A = 65 \, km/h$.
The velocity of object $B$ is $v_B = 80 \, km/h$.
Since both objects are moving in the same direction,the relative velocity of $B$ with respect to $A$ is given by the formula:
$v_{BA} = v_B - v_A$
Substituting the given values:
$v_{BA} = 80 \, km/h - 65 \, km/h = 15 \, km/h$.
Therefore,the relative velocity of $B$ with respect to $A$ is $15 \, km/h$.

(C) Let the man swim at an angle $\theta$ with the downstream direction. The velocity components are $v_x = u + v \cos \theta$ and $v_y = v \sin \theta$.
The time taken to cross the river is $t = \frac{d}{v \sin \theta}$.
The downstream drift is $x = v_x t = (u + v \cos \theta) \frac{d}{v \sin \theta} = \frac{d}{v} (u \csc \theta + v \cot \theta)$.
For maximum time,$\sin \theta$ must be minimum. Since $\theta$ is the angle with the downstream,the range is $0 < \theta < \pi$. The minimum $\sin \theta$ occurs as $\theta \to 0$ or $\theta \to \pi$,but for crossing,$\theta$ must be in $(0, \pi)$. As $\theta \to 0$,$x \to \infty$. Thus,option $(a)$ is incorrect.
To minimize $x$,we differentiate $x$ with respect to $\theta$ and set it to zero: $\frac{dx}{d\theta} = \frac{d}{v} (-u \csc \theta \cot \theta - v \csc^2 \theta) = 0$.
This implies $u \cos \theta + v = 0$,so $\cos \theta = -v/u$.
Since $\theta$ is the angle with the flow,the angle with the normal to the flow is $\alpha = \theta - \pi/2$. Using $\cos \theta = -v/u$,we find $\sin \alpha = v/u$,so $\alpha = \sin^{-1}(v/u)$.
The angle with the flow is $\theta = \pi/2 + \sin^{-1}(v/u)$. Thus,option $(c)$ is correct.

(B) Let $\vec{v}_m$ be the velocity of the bus (man) moving from west to east.
Let $\vec{v}_r$ be the actual velocity of the rain.
Let $\vec{v}_{rm}$ be the velocity of the rain with respect to the man in the bus.
We are given that $\vec{v}_{rm}$ is vertical (downwards).
By the definition of relative velocity,$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$,which implies $\vec{v}_r = \vec{v}_{rm} + \vec{v}_m$.
Since $\vec{v}_{rm}$ is vertical and $\vec{v}_m$ is directed from west to east,the resultant vector $\vec{v}_r$ must have a horizontal component directed from west to east.
Therefore,to a stationary observer on the ground,the rain appears to fall at an angle,moving from west to east.

(B) The boat covers a total distance of $16 \, km$ ($8 \, km$ forward and $8 \, km$ back) in still water in $2 \, h$.
Therefore,the velocity of the boat in still water is $v_B = \frac{16 \, km}{2 \, h} = 8 \, km/h$.
Given the velocity of water is $v_w = 4 \, km/h$.
Time taken to travel $8 \, km$ upstream: $t_1 = \frac{8}{v_B - v_w} = \frac{8}{8 - 4} = \frac{8}{4} = 2 \, h$.
Time taken to travel $8 \, km$ downstream: $t_2 = \frac{8}{v_B + v_w} = \frac{8}{8 + 4} = \frac{8}{12} = \frac{2}{3} \, h = 40 \, min$.
Total time taken = $t_1 + t_2 = 2 \, h + 40 \, min = 2 \, h \, 40 \, min$.

(D) The train is moving towards the west with a velocity $v_t = 10 \, m/s$ (west).
The bird is flying towards the east with a velocity $v_b = 5 \, m/s$ (east).
Since they are moving in opposite directions,their relative velocity $v_{rel}$ is the sum of their individual speeds:
$v_{rel} = v_t + v_b = 10 \, m/s + 5 \, m/s = 15 \, m/s$.
The length of the train to be covered by the bird is $L = 120 \, m$.
The time taken $t$ to cross the train is given by the formula:
$t = \frac{L}{v_{rel}} = \frac{120 \, m}{15 \, m/s} = 8 \, sec$.
Therefore,the time taken by the bird to cross the train is $8 \, sec$.

(B) Let $\overrightarrow{v_b}$ be the velocity of the boat relative to the water and $\overrightarrow{v_r}$ be the velocity of the river water relative to the ground.
The resultant velocity of the boat relative to the ground is given by $\overrightarrow{v_{bg}} = \overrightarrow{v_b} + \overrightarrow{v_r}$.
Since the boat crosses the river perpendicular to the flow,the vectors $\overrightarrow{v_b}$ and $\overrightarrow{v_r}$ are perpendicular to each other.
Therefore,the magnitude of the resultant velocity is $v_{bg} = \sqrt{v_b^2 + v_r^2}$.
Given $v_{bg} = 10 \, km/h$ and $v_b = 8 \, km/h$,we have:
$10 = \sqrt{8^2 + v_r^2}$
$100 = 64 + v_r^2$
$v_r^2 = 100 - 64 = 36$
$v_r = 6 \, km/h$.

(B) The shortest path to cross a river is the straight line perpendicular to the river flow.
Let $v_b = 5 \; km/hr$ be the velocity of the boat and $u$ be the velocity of the river stream.
The resultant velocity $v_r$ of the boat relative to the ground,when crossing the river along the shortest path,is given by $v_r = \sqrt{v_b^2 - u^2}$.
Given width $d = 1 \; km$ and time $t = 15 \; min = 0.25 \; hr = \frac{1}{4} \; hr$.
The resultant velocity is $v_r = \frac{d}{t} = \frac{1}{1/4} = 4 \; km/hr$.
Substituting the values: $4 = \sqrt{5^2 - u^2}$.
Squaring both sides: $16 = 25 - u^2$.
$u^2 = 25 - 16 = 9$.
$u = 3 \; km/hr$.

(D) Let $v_r$ be the speed of the river current and $v_m$ be the speed of the man in still water.
Given that the man crosses the river perpendicular to the current,his resultant velocity $v_{res}$ is given by $v_{res} = \sqrt{v_m^2 - v_r^2}$.
The width of the river is $d = 320 \ m$ and the time taken is $t = 4 \ min$.
Thus,the resultant velocity is $v_{res} = \frac{d}{t} = \frac{320}{4} = 80 \ m/min$.
We are given $v_m = \frac{5}{3} v_r$.
Substituting this into the velocity equation: $80 = \sqrt{(\frac{5}{3} v_r)^2 - v_r^2}$.
Squaring both sides: $80^2 = \frac{25}{9} v_r^2 - v_r^2$.
$6400 = \frac{16}{9} v_r^2$.
$v_r^2 = \frac{6400 \times 9}{16} = 400 \times 9 = 3600$.
$v_r = \sqrt{3600} = 60 \ m/min$.

(B) The total distance to be covered for the trains to completely cross each other is the sum of their lengths: $D = 50 \, m + 50 \, m = 100 \, m$.
Since the trains are moving in opposite directions,their relative velocity is the sum of their individual velocities: $v_{rel} = 10 \, m/s + 15 \, m/s = 25 \, m/s$.
The time taken to cross is given by the formula: $t = \frac{D}{v_{rel}}$.
Substituting the values: $t = \frac{100 \, m}{25 \, m/s} = 4 \, s$.

(A) First,convert all velocities to $m/s$:
Velocity of police jeep $(v_p)$ $= 45 \, km/h = 45 \times \frac{5}{18} \, m/s = 12.5 \, m/s$.
Velocity of thief's jeep $(v_t)$ $= 153 \, km/h = 153 \times \frac{5}{18} \, m/s = 42.5 \, m/s$.
Muzzle velocity of the bullet $(v_b)$ $= 180 \, m/s$.
The velocity of the bullet with respect to the ground is $v_{bg} = v_b + v_p = 180 + 12.5 = 192.5 \, m/s$.
The velocity of the bullet with respect to the thief's car is $v_{bt} = v_{bg} - v_t$.
$v_{bt} = 192.5 \, m/s - 42.5 \, m/s = 150 \, m/s$.

(C) Let $\vec{v}_b$ be the velocity of the boat relative to the water,and $\vec{v}_r$ be the velocity of the river relative to the ground.
The resultant velocity of the boat relative to the ground is $\vec{v} = \vec{v}_b + \vec{v}_r$.
From the given diagram,the velocity of the boat relative to the water is represented by $\vec{AB} = 8 \, km/hr$ (directed perpendicular to the river flow).
The resultant velocity of the boat relative to the ground is represented by $\vec{AC} = 10 \, km/hr$.
The velocity of the river is represented by $\vec{BC}$.
Since $\triangle ABC$ is a right-angled triangle with $\angle ABC = 90^\circ$,we have:
$AC^2 = AB^2 + BC^2$
$BC = \sqrt{AC^2 - AB^2}$
$BC = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \, km/hr$.

(D) The length of the train is $L = 150 \ m$.
The velocity of the train is $v_t = 10 \ m/s$ (towards North).
The velocity of the parrot is $v_p = 5 \ m/s$ (towards South).
Since they are moving in opposite directions,the relative velocity of the parrot with respect to the train is $v_{rel} = v_t + v_p = 10 + 5 = 15 \ m/s$.
The time taken by the parrot to cross the train is $t = \frac{L}{v_{rel}}$.
Substituting the values,$t = \frac{150}{15} = 10 \ s$.

(B) The relative velocity of the boat with respect to the water is given by the formula:
$\vec{v}_{bw} = \vec{v}_b - \vec{v}_w$
Given:
$\vec{v}_b = 3\hat i + 4\hat j$
$\vec{v}_w = -3\hat i - 4\hat j$
Substituting these values into the formula:
$\vec{v}_{bw} = (3\hat i + 4\hat j) - (-3\hat i - 4\hat j)$
$\vec{v}_{bw} = 3\hat i + 4\hat j + 3\hat i + 4\hat j$
$\vec{v}_{bw} = 6\hat i + 8\hat j$
Thus,the relative velocity of the boat with respect to the water is $6\hat i + 8\hat j$.

(D) Let the speed of the boat relative to water be $v_b = 5 \, km/h$ and the speed of the river be $v_r = 3 \, km/h$. The width of the river is $d = 1 \, km$.
To minimize the time for a round trip,the boat must cross the river perpendicular to the flow of the river.
The effective velocity of the boat across the river is $v_{eff} = \sqrt{v_b^2 - v_r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, km/h$.
The time taken to cross the river once is $t = \frac{d}{v_{eff}} = \frac{1 \, km}{4 \, km/h} = 0.25 \, h$.
For a round trip,the boat must cross the river and return,so the total time is $T = 2 \times t = 2 \times 0.25 \, h = 0.5 \, h$.
Converting to minutes,$T = 0.5 \times 60 \, min = 30 \, min$.

(C) For the shortest possible path,the resultant velocity of the man must be perpendicular to the river flow (i.e.,directed due South).
Let $v_r = 5 \, m/min$ be the velocity of the river and $v_m = 10 \, m/min$ be the velocity of the man in still water.
Let $\theta$ be the angle the man makes with the line perpendicular to the river flow (towards the upstream direction).
From the vector triangle,$\sin \theta = \frac{v_r}{v_m} = \frac{5}{10} = \frac{1}{2}$.
Therefore,$\theta = 30^\circ$.
The angle with the downstream direction is $90^\circ + \theta = 90^\circ + 30^\circ = 120^\circ$.

(B) Let the speed of the train be $v$ and the speed of the car be $v$.
Velocity of the train is $\vec{v_t} = v \hat{i}$.
Velocity of the car is $\vec{v_c} = v \hat{j}$.
The velocity of the car with respect to the train is given by $\vec{v_{ct}} = \vec{v_c} - \vec{v_t}$.
Substituting the values,we get $\vec{v_{ct}} = v \hat{j} - v \hat{i} = v(-\hat{i} + \hat{j})$.
The vector $(-\hat{i} + \hat{j})$ points in the direction between the West $(-\hat{i})$ and North $(+\hat{j})$ directions.
Therefore,the observed direction of the car to a passenger in the train is West-north.

(C) Since the man is inside the train,he and the coin share the same initial horizontal velocity as the train.
Because the horizontal velocity is the same for both the coin and the observer (the man),the relative horizontal velocity between them is zero.
Therefore,the coin will only experience vertical acceleration due to gravity relative to the man.
As a result,the path of the coin as observed by the man will be a vertical straight line.

(D) For two particles to collide,their positions at time $t$ must be equal. Let $\overrightarrow{r_1}(t)$ and $\overrightarrow{r_2}(t)$ be the positions of the particles at time $t$.
$\overrightarrow{r_1}(t) = \overrightarrow{r_1} + \overrightarrow{v_1}t = (3\hat{i} + 5\hat{j}) + (4\hat{i} + 3\hat{j})t = (3+4t)\hat{i} + (5+3t)\hat{j}$
$\overrightarrow{r_2}(t) = \overrightarrow{r_2} + \overrightarrow{v_2}t = (-5\hat{i} - 3\hat{j}) + (\alpha\hat{i} + 7\hat{j})t = (-5+\alpha t)\hat{i} + (-3+7t)\hat{j}$
At collision,$\overrightarrow{r_1}(t) = \overrightarrow{r_2}(t)$ at $t = 2 \text{ s}$.
Equating the $\hat{i}$ components:
$3 + 4(2) = -5 + \alpha(2)$
$3 + 8 = -5 + 2\alpha$
$11 = -5 + 2\alpha$
$16 = 2\alpha$
$\alpha = 8$
Equating the $\hat{j}$ components to verify:
$5 + 3(2) = -3 + 7(2)$
$5 + 6 = -3 + 14$
$11 = 11$. This confirms the result.

(B) Let the horizontal distance covered by particle $A$ be $x_1$ and by particle $B$ be $x_2$.
Particle $A$ moves with velocity $v_A = u/\sqrt{3}$ at an angle of $30^\circ$ with the horizontal. Its horizontal component of velocity is $v_{Ax} = (u/\sqrt{3}) \cos 30^\circ = (u/\sqrt{3}) \times (\sqrt{3}/2) = u/2$.
Particle $B$ moves with velocity $v_B = u$ at an angle of $60^\circ$ with the horizontal. Its horizontal component of velocity is $v_{Bx} = u \cos 60^\circ = u \times (1/2) = u/2$.
Since both particles are moving towards each other,their relative horizontal velocity is $v_{rel} = v_{Ax} + v_{Bx} = u/2 + u/2 = u$.
The initial horizontal separation between them is $x$.
Time taken for the horizontal separation to become zero is $t = \text{Distance} / \text{Relative Velocity} = x / u$.

(B) Let $d$ be the distance of the escalator.
Velocity of Preeti on the stationary escalator is $v_1 = \frac{d}{t_1}$.
Velocity of the moving escalator is $v_2 = \frac{d}{t_2}$.
When Preeti walks on the moving escalator,her net velocity with respect to the ground is $v = v_1 + v_2$.
$v = \frac{d}{t_1} + \frac{d}{t_2} = d \left( \frac{t_1 + t_2}{t_1 t_2} \right)$.
The time $t$ taken to cover the distance $d$ with net velocity $v$ is:
$t = \frac{d}{v} = \frac{d}{d \left( \frac{t_1 + t_2}{t_1 t_2} \right)} = \frac{t_1 t_2}{t_1 + t_2}$.

(B) Let the initial position of ship $A$ be at the origin $(0, 0)$ moving along the negative $x$-axis. Its position at time $t$ is $\vec{r}_A = (-10t, 0)$.
Ship $B$ is initially at $(0, -100)$ moving along the positive $y$-axis. Its position at time $t$ is $\vec{r}_B = (0, -100 + 10t)$.
The relative position vector is $\vec{r}_{BA} = \vec{r}_B - \vec{r}_A = (10t, -100 + 10t)$.
The square of the distance $D^2 = (10t)^2 + (-100 + 10t)^2 = 100t^2 + 10000 - 2000t + 100t^2 = 200t^2 - 2000t + 10000$.
To find the shortest distance,differentiate $D^2$ with respect to $t$ and set it to zero:
$\frac{d(D^2)}{dt} = 400t - 2000 = 0$.
$400t = 2000 \Rightarrow t = 5 \, hr$.
Alternatively,using relative velocity: $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = (0, 10) - (-10, 0) = (10, 10) \, km \, h^{-1}$.
The magnitude of relative velocity is $|\vec{v}_{BA}| = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, km \, h^{-1}$.
The shortest distance occurs when the relative position vector is perpendicular to the relative velocity vector. From the geometry,the time taken is $t = 5 \, hr$.

(A) Let the particles $A$ and $B$ collide at time $t$. For their collision,the position vectors of both particles must be the same at time $t$,i.e.,
$\vec{r}_1 + \vec{v}_1 t = \vec{r}_2 + \vec{v}_2 t$
$\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1) t$ ... $(i)$
Taking the magnitude of both sides:
$|\vec{r}_1 - \vec{r}_2| = |\vec{v}_2 - \vec{v}_1| t$
$t = \frac{|\vec{r}_1 - \vec{r}_2|}{|\vec{v}_2 - \vec{v}_1|}$
Substituting this value of $t$ into equation $(i)$:
$\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1) \frac{|\vec{r}_1 - \vec{r}_2|}{|\vec{v}_2 - \vec{v}_1|}$
Rearranging the terms,we get:
$\frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \frac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$

(B) Let the breadth of the lake be $l$ and the velocity of the boat be $v_b$.
On a quiet day,the time taken in going and coming back is:
$t_Q = \frac{l}{v_b} + \frac{l}{v_b} = \frac{2l}{v_b}$ .....$(i)$
Now,if $v_a$ is the velocity of the air current,the time taken in going across the lake is:
$t_1 = \frac{l}{v_b + v_a}$ [As the current helps the motion]
And the time taken in coming back is:
$t_2 = \frac{l}{v_b - v_a}$ [As the current opposes the motion]
So,the total time on a rough day is:
$t_R = t_1 + t_2 = \frac{l(v_b - v_a) + l(v_b + v_a)}{v_b^2 - v_a^2} = \frac{2lv_b}{v_b^2 - v_a^2} = \frac{2l}{v_b[1 - (v_a/v_b)^2]}$ .....(ii)
Comparing equations $(i)$ and (ii):
$\frac{t_R}{t_Q} = \frac{1}{1 - (v_a/v_b)^2}$. Since $1 - (v_a/v_b)^2 < 1$,it follows that $\frac{t_R}{t_Q} > 1$,which means $t_R > t_Q$.
Therefore,the time taken to complete the journey on a quiet day is less than that on a rough day.

(A) Let the velocity of the car be $\vec{v}_C = 25\hat{i}\, km/hr$ (eastward).
Let the velocity of the train be $\vec{v}_T = v_x\hat{i} + v_y\hat{j}$.
The velocity of the train relative to the car is $\vec{v}_{TC} = \vec{v}_T - \vec{v}_C = (v_x - 25)\hat{i} + v_y\hat{j}$.
Given that the train appears to move north,the horizontal component of the relative velocity must be zero: $v_x - 25 = 0$,so $v_x = 25\, km/hr$.
The magnitude of the relative velocity is given as $25\sqrt{3}\, km/hr$,which is the vertical component: $v_y = 25\sqrt{3}\, km/hr$.
The actual velocity of the train is $\vec{v}_T = 25\hat{i} + 25\sqrt{3}\hat{j}$.
The magnitude is $|\vec{v}_T| = \sqrt{25^2 + (25\sqrt{3})^2} = \sqrt{625 + 1875} = \sqrt{2500} = 50\, km/hr$.

(B) Let the width of the river be $d$.
For the shortest time,the swimmer swims perpendicular to the river flow. The time taken is $t = \frac{d}{v}$.
For the shortest distance,the swimmer must swim at an angle $\theta$ with the upstream such that the resultant velocity is perpendicular to the river bank. The component of velocity perpendicular to the bank is $v \sin \theta$ (where $\theta$ is the angle with the upstream). Thus,the time taken is $t' = \frac{d}{v \sin \theta}$.
The ratio of the time taken for the shortest time to the time taken for the shortest distance is $\frac{t}{t'} = \frac{d/v}{d/(v \sin \theta)} = \sin \theta$.

(B) Let the length of the escalator be $x$.
Speed of the escalator,$v_e = \frac{x}{1} = x \text{ m/min}$.
Speed of the man walking on a stationary escalator,$v_m = \frac{x}{3} \text{ m/min}$.
When the man walks up the moving escalator,his effective speed is $v_{eff} = v_e + v_m = x + \frac{x}{3} = \frac{4x}{3} \text{ m/min}$.
The time taken to reach the top is $t = \frac{x}{v_{eff}} = \frac{x}{4x/3} = \frac{3}{4} \text{ minutes}$.
Converting to seconds: $t = \frac{3}{4} \times 60 \text{ s} = 45 \text{ s}$.

(D) Let the intersection point $O$ be the origin $(0,0)$. Let one particle $Q$ move along the $x$-axis towards $O$ and the other particle $P$ move along a line at $60^{\circ}$ to the $x$-axis towards $O$.
The velocities are $\vec{v}_Q = -20 \hat{i} \, cm/s$ and $\vec{v}_P = -20 \cos 60^{\circ} \hat{i} - 20 \sin 60^{\circ} \hat{j} = -10 \hat{i} - 10\sqrt{3} \hat{j} \, cm/s$.
The relative velocity of $Q$ with respect to $P$ is $\vec{v}_{QP} = \vec{v}_Q - \vec{v}_P = (-20 - (-10)) \hat{i} - (-10\sqrt{3}) \hat{j} = -10 \hat{i} + 10\sqrt{3} \hat{j} \, cm/s$.
The initial positions are $\vec{r}_Q = 400 \hat{i} \, cm$ and $\vec{r}_P = 300 \cos 60^{\circ} \hat{i} + 300 \sin 60^{\circ} \hat{j} = 150 \hat{i} + 150\sqrt{3} \hat{j} \, cm$.
The relative position is $\vec{r}_{QP} = \vec{r}_Q - \vec{r}_P = (400 - 150) \hat{i} - 150\sqrt{3} \hat{j} = 250 \hat{i} - 150\sqrt{3} \hat{j} \, cm$.
The shortest distance is given by $d = \frac{|\vec{r}_{QP} \times \vec{v}_{QP}|}{|\vec{v}_{QP}|}$.
$|\vec{v}_{QP}| = \sqrt{(-10)^2 + (10\sqrt{3})^2} = \sqrt{100 + 300} = 20 \, cm/s$.
The cross product $\vec{r}_{QP} \times \vec{v}_{QP} = (250 \hat{i} - 150\sqrt{3} \hat{j}) \times (-10 \hat{i} + 10\sqrt{3} \hat{j}) = (250 \times 10\sqrt{3} - (-150\sqrt{3}) \times (-10)) \hat{k} = (2500\sqrt{3} - 1500\sqrt{3}) \hat{k} = 1000\sqrt{3} \hat{k}$.
The magnitude is $1000\sqrt{3}$.
Therefore,$d = \frac{1000\sqrt{3}}{20} = 50\sqrt{3} \, cm$.

(B) Let $V_{SR} = 5 \text{ km/hr}$ be the velocity of the swimmer with respect to the river.
Let $V_R = 4 \text{ km/hr}$ be the velocity of the river with respect to the ground.
The angle with the river flow is $127^{\circ}$. The velocity components of the swimmer relative to the ground are:
$V_{Sx} = V_R + V_{SR} \cos(127^{\circ}) = 4 + 5(-\cos 53^{\circ}) = 4 - 5(0.6) = 4 - 3 = 1 \text{ km/hr}$.
$V_{Sy} = V_{SR} \sin(127^{\circ}) = 5 \sin(53^{\circ}) = 5(0.8) = 4 \text{ km/hr}$.
The time taken to cross the river of width $d = 0.2 \text{ km}$ is $t_1 = \frac{d}{V_{Sy}} = \frac{0.2}{4} = 0.05 \text{ hr} = 3 \text{ minutes}$.
The horizontal distance covered downstream is $CB = V_{Sx} \times t_1 = 1 \text{ km/hr} \times 0.05 \text{ hr} = 0.05 \text{ km} = 50 \text{ m}$.
Time taken to walk distance $CB$ at $3 \text{ km/hr}$ is $t_2 = \frac{0.05 \text{ km}}{3 \text{ km/hr}} = \frac{1}{60} \text{ hr} = 1 \text{ minute}$.
Total time $= t_1 + t_2 = 3 \text{ min} + 1 \text{ min} = 4 \text{ minutes}$.

(B) The shortest path to cross a river is the straight line perpendicular to the river banks. Let $v_b = 5 \, km/hr$ be the speed of the boat in still water and $v_r$ be the speed of the river.
The resultant velocity $v_{res}$ along the shortest path is given by:
$v_{res} = \frac{\text{width of river}}{\text{time taken}} = \frac{1 \, km}{15 \, min} = \frac{1 \, km}{15/60 \, hr} = 4 \, km/hr$.
In the vector triangle for crossing the river along the shortest path,the boat's velocity in still water $(v_b)$ acts as the hypotenuse,the resultant velocity $(v_{res})$ acts as one side,and the river's velocity $(v_r)$ acts as the other side.
According to the Pythagorean theorem:
$v_b^2 = v_{res}^2 + v_r^2$
$5^2 = 4^2 + v_r^2$
$25 = 16 + v_r^2$
$v_r^2 = 25 - 16 = 9$
$v_r = 3 \, km/hr$.

(C) The velocity of the car is $\vec{v}_c = 20 \hat{j} \, km/hr$ (North).
The velocity of the wind is $\vec{v}_w = 20 \hat{i} \, km/hr$ (East).
The flag points in the direction of the relative velocity of the wind with respect to the car,which is $\vec{v}_{wc} = \vec{v}_w - \vec{v}_c$.
$\vec{v}_{wc} = 20 \hat{i} - 20 \hat{j}$.
The direction of this vector is given by $\tan \theta = \frac{|v_y|}{|v_x|} = \frac{20}{20} = 1$,so $\theta = 45^\circ$.
Since the $x$-component is positive (East) and the $y$-component is negative (South),the resultant direction is South-East.

(B) Let $v$ be the velocity of the man in still water and $u = 5\, m/s$ be the velocity of the river flow.
To reach a point directly across,the man must swim at an angle against the current such that the resultant velocity is perpendicular to the river bank.
The effective velocity of the man perpendicular to the bank is $v_{eff} = \sqrt{v^2 - u^2}$.
The time taken to cross the river is given by $t = \frac{d}{v_{eff}}$,where $d = 60\, m$ is the width of the river.
Substituting the given values: $5 = \frac{60}{\sqrt{v^2 - 5^2}}$.
$\sqrt{v^2 - 25} = \frac{60}{5} = 12$.
Squaring both sides: $v^2 - 25 = 144$.
$v^2 = 144 + 25 = 169$.
$v = \sqrt{169} = 13\, m/s$.

(B) Let $v_s$ be the speed of the swimmer in still water and $v_r$ be the speed of the river.
When the swimmer moves downstream,their effective speed is $(v_s + v_r)$.
When the swimmer moves upstream,their effective speed is $(v_s - v_r)$.
The float moves with the speed of the river,$v_r$.
Let $t_1$ be the time taken to travel distance $D$ downstream,and $t_2$ be the time taken to travel back to the float.
Time $t_1 = D / (v_s + v_r)$.
At this point,the float has moved a distance $x = v_r t_1 = v_r D / (v_s + v_r)$.
After turning back,the swimmer meets the float at a distance $D/2$ from $M$. This means the swimmer covers a distance $D - D/2 = D/2$ upstream.
The float covers a distance $D/2 - x$ in the same time $t_2$.
$t_2 = (D/2) / (v_s - v_r) = (D/2 - x) / v_r$.
Substituting $x$,we get $t_2 = (D/2) / (v_s - v_r) = (D/2 - v_r D / (v_s + v_r)) / v_r$.
Dividing by $D$ and simplifying: $1 / (2(v_s - v_r)) = (1/2 - v_r / (v_s + v_r)) / v_r$.
$1 / (2(v_s - v_r)) = (v_s + v_r - 2v_r) / (2v_r(v_s + v_r)) = (v_s - v_r) / (2v_r(v_s + v_r))$.
$(v_s + v_r) = (v_s - v_r)^2 / v_r$.
Let $k = v_s / v_r$. Then $k + 1 = (k - 1)^2 = k^2 - 2k + 1$.
$k^2 - 3k = 0$. Since $k \neq 0$,$k = 3$.

(A) Let the velocity of the car be $\vec{v}_c = 2\hat{i}$ and the velocity of the rain drops be $\vec{v}_r = -6\hat{j}$.
The velocity of the rain drops relative to the car is $\vec{v}_{rc} = \vec{v}_r - \vec{v}_c = -2\hat{i} - 6\hat{j}$.
For the rain drops to strike the windscreen perpendicularly,the windscreen must be perpendicular to the relative velocity vector $\vec{v}_{rc}$.
The angle $\alpha$ that the relative velocity vector $\vec{v}_{rc}$ makes with the vertical ($-y$ axis) is given by $\tan \alpha = \frac{|v_x|}{|v_y|} = \frac{2}{6} = \frac{1}{3}$.
However,the windscreen is inclined at an angle $\alpha$ with the vertical such that its normal is parallel to the relative velocity vector. The angle of the windscreen with the vertical is the same as the angle the relative velocity vector makes with the horizontal,which is $\theta = \tan^{-1}(\frac{6}{2}) = \tan^{-1}(3)$.

(B) The horizontal component of the rain's velocity is equal to the velocity of the wind,which is $2 \, m/s$ in the north direction.
Let $\vec{v}_r$ be the velocity of the rain and $\vec{v}_w$ be the velocity of the wind. The horizontal component of the rain is $\vec{v}_{r,h} = \vec{v}_w = 2 \, m/s$ (North).
For the cyclist,the rain appears to fall vertically if the relative horizontal velocity of the rain with respect to the cyclist is zero.
Let $\vec{v}_c$ be the velocity of the cyclist. The relative velocity of the rain with respect to the cyclist is $\vec{v}_{r/c} = \vec{v}_r - \vec{v}_c$.
For the horizontal component to be zero,the cyclist must have the same horizontal velocity as the rain.
Therefore,$\vec{v}_c = 2 \, m/s$ in the north direction.

(D) Let the observer move with velocity $v$ along the line joining two stationary objects $A$ and $B$.
Since the objects are stationary in the ground frame,their velocities are $v_A = 0$ and $v_B = 0$.
The velocity of the objects relative to the observer is given by $v_{\text{rel}} = v_{\text{object}} - v_{\text{observer}}$.
For both objects,$v_{\text{rel}} = 0 - v = -v$.
This means both objects appear to move with the same speed $v$ in the direction opposite to the observer's motion.
Since both have the same velocity vector ($-v$ relative to the observer),all the given statements are correct.

(A) Let the velocity of the river be $v_r = 5 \ km/hr$ and the velocity of the boat in still water be $v_{br} = 3 \ km/hr$.
Since $v_{br} < v_r$,the boat cannot reach the point directly opposite to $A$ (point $B$).
To minimize the drift,the boat must be steered at an angle such that its resultant velocity is perpendicular to the river flow.
However,the problem states the boat reaches point $C$ while maintaining a straight path.
Given the standard interpretation of such problems where the boat aims for the shortest path but is carried downstream,the drift $BC$ is given by $BC = (v_r - v_{br} \sin \theta) \times t$,where $t = b / (v_{br} \cos \theta)$.
For the minimum distance path,the boat is steered at an angle $\theta$ such that the component of velocity perpendicular to the river is maximized.
Given the parameters,the resultant velocity $v_b$ makes an angle with the bank.
The distance $AC = \sqrt{b^2 + BC^2}$.
Based on the provided options and the geometry,the correct distance is $500 \ m$.

(A) Let the tower be at the origin $(0, 80)$. The initial velocity of shell $1$ is $\vec{v}_1 = 200(\cos 37^o \hat{i} + \sin 37^o \hat{j}) = 200(0.8 \hat{i} + 0.6 \hat{j}) = 160 \hat{i} + 120 \hat{j}\ m/s$.
The jeep moves with $\vec{v}_J = 10 \hat{i}\ m/s$. The velocity of shell $2$ relative to the jeep is $\vec{v}_{2J} = 250(\cos 53^o \hat{i} + \sin 53^o \hat{j}) = 250(0.6 \hat{i} + 0.8 \hat{j}) = 150 \hat{i} + 200 \hat{j}\ m/s$.
The absolute velocity of shell $2$ is $\vec{v}_2 = \vec{v}_{2J} + \vec{v}_J = (150 \hat{i} + 200 \hat{j}) + 10 \hat{i} = 160 \hat{i} + 200 \hat{j}\ m/s$.
Since both shells are under gravity,their relative acceleration is $\vec{a}_{21} = \vec{a}_2 - \vec{a}_1 = (-g \hat{j}) - (-g \hat{j}) = 0$. Thus,the relative motion is uniform.
The relative velocity is $\vec{v}_{21} = \vec{v}_2 - \vec{v}_1 = (160 \hat{i} + 200 \hat{j}) - (160 \hat{i} + 120 \hat{j}) = 80 \hat{j}\ m/s$.
The initial relative position is $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1 = (0, 0) - (0, 80) = -80 \hat{j}\ m$.
The relative position at time $t$ is $\vec{r}(t) = \vec{r}_{21} + \vec{v}_{21} t = -80 \hat{j} + 80t \hat{j} = 80(t-1) \hat{j}$.
The shells are closest when the magnitude of the relative position vector is minimum,which occurs at $t = 1\ s$,where the relative distance is $0$.

(A) Let the velocity of the river be $\vec{u}$ and the velocity of the swimmer with respect to the water be $\vec{v}$.
In the frame of reference of the river,the float is at rest.
The swimmer moves away from the float with a speed '$v$' relative to the water (and thus relative to the float) for a distance '$l$'.
The time taken to move away is $t_1 = \frac{l}{v}$.
Then,the swimmer turns back and moves towards the float with a speed '$v$' relative to the water (and thus relative to the float) for the same distance '$l$'.
The time taken to return is $t_2 = \frac{l}{v}$.
The total time taken by the swimmer in this process is $T = t_1 + t_2 = \frac{l}{v} + \frac{l}{v} = \frac{2l}{v}$.

(B) Let the position of $Q$ be at the origin $(0, 0)$ and $P$ be at $(0, 10)$.
The velocity of $P$ is $\vec{v}_P = -10 \hat{i} \ kmph$.
The velocity of $Q$ is $\vec{v}_Q = 10 \hat{j} \ kmph$.
The relative velocity of $P$ with respect to $Q$ is $\vec{v}_{PQ} = \vec{v}_P - \vec{v}_Q = -10 \hat{i} - 10 \hat{j} \ kmph$.
The magnitude of relative velocity is $|\vec{v}_{PQ}| = \sqrt{(-10)^2 + (-10)^2} = 10\sqrt{2} \ kmph$.
The angle made by the relative velocity vector with the $y$-axis is $45^\circ$.
The minimum separation occurs when the position vector of $P$ relative to $Q$ is perpendicular to the relative velocity vector $\vec{v}_{PQ}$.
The time taken is $t = \frac{d \cos 45^\circ}{|\vec{v}_{PQ}|} = \frac{10 \times \frac{1}{\sqrt{2}}}{10\sqrt{2}} = \frac{1}{2} = 0.5 \ hr$.

(A) Given:
Velocity of block $A$,$\vec{v}_A = 2 \ m/s$ at $75^\circ$ below the horizontal.
Velocity of belt portion $CD$ relative to $A$,$\vec{v}_{CD/A} = 2 \ m/s$ at $\theta = 15^\circ$ with the horizontal.
We know that $\vec{v}_{CD} = \vec{v}_{CD/A} + \vec{v}_A$.
The angle between $\vec{v}_A$ and $\vec{v}_{CD/A}$ is $\alpha = 75^\circ - 15^\circ = 60^\circ$.
Using the law of cosines for vector addition:
$v_{CD}^2 = v_A^2 + v_{CD/A}^2 + 2 v_A v_{CD/A} \cos(60^\circ)$
$v_{CD}^2 = 2^2 + 2^2 + 2(2)(2)(0.5)$
$v_{CD}^2 = 4 + 4 + 4 = 12$
$v_{CD} = \sqrt{12} = 2\sqrt{3} \ m/s$.