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

(D) For two particles to collide,their positions at time $t$ must be equal: $r_1 + v_1 t = r_2 + v_2 t$.
Rearranging the equation: $r_1 - r_2 = (v_2 - v_1) t$.
Given $r_1 = (3 \hat{i} + 5 \hat{j})$ and $r_2 = (-5 \hat{i} - 3 \hat{j})$,the relative position vector is $r_1 - r_2 = (3 - (-5)) \hat{i} + (5 - (-3)) \hat{j} = (8 \hat{i} + 8 \hat{j}) \text{ m}$.
The relative velocity is $v_2 - v_1 = (a \hat{i} + 7 \hat{j}) - (4 \hat{i} + 3 \hat{j}) = (a - 4) \hat{i} + 4 \hat{j}$.
Substituting these into the collision condition with $t = 2 \text{ s}$:
$8 \hat{i} + 8 \hat{j} = ((a - 4) \hat{i} + 4 \hat{j}) \times 2$.
Dividing by $2$: $4 \hat{i} + 4 \hat{j} = (a - 4) \hat{i} + 4 \hat{j}$.
Comparing the coefficients of $\hat{i}$: $4 = a - 4$,which gives $a = 8$.

(B) Let the speed of the boat in still water be $v_b$ and the speed of the stream be $v_s$.
Downstream speed is $v_b + v_s = \frac{16 \, km}{2 \, h} = 8 \, km/h$.
Upstream speed is $v_b - v_s = \frac{16 \, km}{4 \, h} = 4 \, km/h$.
Adding the two equations: $(v_b + v_s) + (v_b - v_s) = 8 + 4$.
$2v_b = 12 \, km/h$.
Therefore, the speed of the boat in still water is $v_b = 6 \, km/h$.

(C) Given: Length of the train, $l = 100 \,m$.
Time taken to cross the object, $t = 7.2 \,s$.
Velocity of the object moving in the opposite direction, $v_o = 5 \,km/h = 5 \times \frac{5}{18} \,m/s = \frac{25}{18} \,m/s \approx 1.39 \,m/s$.
Let the velocity of the train be $v_t = v$.
Since the objects are moving in opposite directions, the relative velocity is $v_{\text{rel}} = v_t + v_o$.
The distance covered to cross the object is equal to the length of the train, $l = v_{\text{rel}} \times t$.
$100 = (v + \frac{25}{18}) \times 7.2$.
Dividing by $7.2$: $\frac{100}{7.2} = v + \frac{25}{18}$.
$\frac{1000}{72} = v + \frac{25}{18} \Rightarrow \frac{125}{9} = v + \frac{25}{18}$.
$v = \frac{250}{18} - \frac{25}{18} = \frac{225}{18} = 12.5 \,m/s$.
Converting to $km/h$: $v = 12.5 \times \frac{18}{5} = 45 \,km/h$.

(C) Let the speed of the bus be $v_B$ and the speed of the cyclist be $v_C = 20 \,km/h$.
Let $d$ be the distance between two consecutive buses,where $d = v_B \times T$.
Case $1$: Bus moving from $A$ to $B$ (same direction as cyclist).
The relative speed of the bus with respect to the cyclist is $(v_B - v_C)$.
The time interval between buses passing the cyclist is $t_1 = 18 \,min = \frac{18}{60} \,h = 0.3 \,h$.
Thus,$d = (v_B - 20) \times 0.3 \dots (i)$.
Case $2$: Bus moving from $B$ to $A$ (opposite direction to cyclist).
The relative speed of the bus with respect to the cyclist is $(v_B + v_C)$.
The time interval between buses passing the cyclist is $t_2 = 6 \,min = \frac{6}{60} \,h = 0.1 \,h$.
Thus,$d = (v_B + 20) \times 0.1 \dots (ii)$.
Equating $(i)$ and $(ii)$:
$(v_B - 20) \times 0.3 = (v_B + 20) \times 0.1$
$3(v_B - 20) = v_B + 20$
$3v_B - 60 = v_B + 20$
$2v_B = 80 \Rightarrow v_B = 40 \,km/h$.
Now,substitute $v_B$ into $d = v_B \times T$ using equation $(ii)$:
$v_B \times T = (v_B + 20) \times 0.1$
$40 \times T = (40 + 20) \times 0.1$
$40T = 60 \times 0.1 = 6$
$T = \frac{6}{40} \,h = \frac{3}{20} \,h$.

(C) Let the direction of the train (North) be positive and the direction of the parrot (South) be negative.
Velocity of the train,$v_t = 10 \,ms^{-1}$.
Velocity of the parrot,$v_p = -5 \,ms^{-1}$.
The relative velocity of the parrot with respect to the train is given by $v_{pt} = v_t - v_p$.
$v_{pt} = 10 - (-5) = 15 \,ms^{-1}$.
The length of the train is $L = 150 \,m$.
The time $t$ taken by the parrot to fly alongside the train is the time taken to cover the length of the train with the relative velocity.
$t = \frac{L}{v_{pt}} = \frac{150}{15} = 10 \,s$.

(A) Let $L$ be the length of the escalator.
Let $v_p$ be the speed of the person walking and $v_e$ be the speed of the escalator.
When the escalator is stalled,the person walks the length $L$ in $t_1 = 90 \ s$. Thus,$v_p = \frac{L}{90}$.
When the person stands on the moving escalator,they reach the top in $t_2 = 60 \ s$. Thus,$v_e = \frac{L}{60}$.
When the person walks up the moving escalator,their effective speed is $v_{eff} = v_p + v_e$.
The time taken $t_3$ is given by $t_3 = \frac{L}{v_p + v_e} = \frac{L}{\frac{L}{90} + \frac{L}{60}}$.
$t_3 = \frac{1}{\frac{1}{90} + \frac{1}{60}} = \frac{90 \times 60}{90 + 60} = \frac{5400}{150} = 36 \ s$.

(D) Let the distance of the escalator be $D$.
Speed of the person walking on the stalled escalator is $v_p = \frac{D}{80}$.
Speed of the moving escalator is $v_e = \frac{D}{20}$.
When the person walks up the moving escalator, his effective speed is $v_{eff} = v_p + v_e$.
$v_{eff} = \frac{D}{80} + \frac{D}{20} = \frac{D + 4D}{80} = \frac{5D}{80} = \frac{D}{16}$.
The time taken $t$ to walk up the moving escalator is $t = \frac{D}{v_{eff}} = \frac{D}{D/16} = 16 \,s$.

(A) The total time taken for the two cars to meet is given by the formula:
$T = \frac{\text{Total distance}}{\text{Relative velocity of the cars}}$
Given, distance $d = 36 \text{ km}$, speed of car $1$ $v_1 = 54 \text{ kmh}^{-1}$, and speed of car $2$ $v_2 = 36 \text{ kmh}^{-1}$.
Since they are moving towards each other, relative velocity $v_{rel} = v_1 + v_2 = 54 + 36 = 90 \text{ kmh}^{-1}$.
$T = \frac{36}{90} = 0.4 \text{ h}$.
The bird flies continuously at a speed of $v_b = 36 \text{ kmh}^{-1}$ for the entire duration $T$.
Total distance covered by the bird $= v_b \times T = 36 \times 0.4 = 14.4 \text{ km}$.
Converting to meters: $14.4 \times 1000 = 14400 \text{ m}$.

(C) The time taken to cross a river of width $d$ is given by $t = \frac{d}{V \cos \theta}$,where $\theta$ is the angle the swimmer makes with the perpendicular to the river flow.
To minimize the time $t$,the denominator $V \cos \theta$ must be maximized.
This occurs when $\cos \theta$ is maximum,which is at $\theta = 0^{\circ}$.
Therefore,the swimmer must swim perpendicular to the flow of the river to cross it in the least time.

(C) Let $v_{CE}$ be the velocity of the car with respect to the Earth,$v_{RE}$ be the velocity of the rain with respect to the Earth,and $v_{RC}$ be the velocity of the rain with respect to the car.
Given: $v_{CE} = 40 \text{ km h}^{-1}$.
The rain falls vertically,so $v_{RE}$ is along the vertical direction.
The trace of the rain on the side window makes an angle $\theta = 30^{\circ}$ with the vertical.
From the vector triangle of relative velocity,we have $\vec{v}_{RC} = \vec{v}_{RE} - \vec{v}_{CE}$.
In the right-angled triangle formed by these vectors,the side representing the car's velocity $v_{CE}$ is opposite to the angle $\theta = 30^{\circ}$.
Therefore,$\sin \theta = \frac{v_{CE}}{v_{RC}}$.
Rearranging for $v_{RC}$,we get $v_{RC} = \frac{v_{CE}}{\sin 30^{\circ}}$.
Substituting the values,$v_{RC} = \frac{40}{0.5} = 80 \text{ km h}^{-1}$.

(A) Let the velocity of cars $A$ and $B$ be $v_A = v_B = 30 \ km/h$ in the positive direction.
Let the velocity of car $C$ be $v_C$ in the opposite direction,so $v_C = -v$ (where $v$ is the speed of car $C$).
The relative velocity of car $C$ with respect to cars $A$ and $B$ is $v_{rel} = v_A - v_C = 30 - (-v) = 30 + v$.
The distance between car $A$ and $B$ is $d = 10 \ km$.
Car $C$ meets car $B$ and then car $A$ after an interval of $t = 8 \ minutes = \frac{8}{60} \ h = \frac{2}{15} \ h$.
Using the formula $d = v_{rel} \times t$:
$10 = (30 + v) \times \frac{2}{15}$
$10 \times \frac{15}{2} = 30 + v$
$75 = 30 + v$
$v = 75 - 30 = 45 \ km/h$.

(C) Let $d$ be the width of the river. The swimmer swims with speed $v$ in still water.
For the shortest time,the swimmer must swim perpendicular to the river flow. The time taken is $t = \frac{d}{v}$.
For the shortest distance,the swimmer must swim at an angle such that the resultant velocity is perpendicular to the river bank. Let the angle with the normal to the stream be $\alpha$. Then $\sin \alpha = \frac{v_{river}}{v} = \frac{v/2}{v} = \frac{1}{2}$,so $\alpha = 30^{\circ}$.
The angle with the upstream is $\theta = 90^{\circ} + \alpha = 120^{\circ}$.
The time taken for the shortest distance is $t^{\prime} = \frac{d}{v \cos \alpha} = \frac{d}{v \sin \theta}$.
The ratio of the time taken is $\frac{t}{t^{\prime}} = \frac{d/v}{d/(v \sin \theta)} = \sin \theta$.

(A) Let the position of ship $A$ at $t=0$ be $(0, 200)$ and ship $B$ be $(0, 0)$.
Velocity of ship $A$ is $\vec{v}_A = -20 \hat{i} \ km h^{-1}$.
Velocity of ship $B$ is $\vec{v}_B = 10 \hat{j} \ km h^{-1}$.
Relative velocity $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = 20 \hat{i} + 10 \hat{j} \ km h^{-1}$.
Relative position $\vec{r}_{BA} = \vec{r}_B - \vec{r}_A = (0 - 0) \hat{i} + (0 - 200) \hat{j} = -200 \hat{j} \ km$.
The distance $d$ at time $t$ is given by $|\vec{r}_{BA} + \vec{v}_{BA} t| = |20t \hat{i} + (10t - 200) \hat{j}|$.
$d^2 = (20t)^2 + (10t - 200)^2 = 400t^2 + 100t^2 - 4000t + 40000 = 500t^2 - 4000t + 40000$.
For minimum distance,$\frac{d(d^2)}{dt} = 1000t - 4000 = 0 \implies t = 4 \ h$.
Shortest distance $d = \sqrt{500(4)^2 - 4000(4) + 40000} = \sqrt{8000 - 16000 + 40000} = \sqrt{32000} = 80 \sqrt{5} \ km$.

(D) Let $\vec{v}_m$ be the velocity of the man,$\vec{v}_r$ be the velocity of the rain,and $\vec{v}_{rm}$ be the velocity of the rain with respect to the man.
Given,$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$.
When the man is running,the rain appears to fall vertically,meaning $\vec{v}_{rm}$ is vertical.
When the man stops,the rain falls at $60^{\circ}$ with the horizontal,meaning $\vec{v}_r$ makes an angle of $60^{\circ}$ with the horizontal.
From the vector triangle,the angle between $\vec{v}_r$ and the vertical is $90^{\circ} - 60^{\circ} = 30^{\circ}$.
In the right-angled triangle formed by $\vec{v}_m$,$\vec{v}_r$,and $\vec{v}_{rm}$:
$\tan(30^{\circ}) = \frac{|\vec{v}_m|}{|\vec{v}_{rm}|}$
$\frac{1}{\sqrt{3}} = \frac{5}{|\vec{v}_{rm}|}$
$|\vec{v}_{rm}| = 5\sqrt{3} \text{ km/h}$.

(D) The total distance to be covered is $2d$ (distance $d$ upstream and $d$ downstream).
In still water,the man swims at speed $2v$. The time taken to cover distance $2d$ is $t_0 = \frac{2d}{2v} = \frac{d}{v}$.
When swimming downstream,the effective speed is $v_{down} = v_m + v_r = 2v + v = 3v$. The time taken is $t_1 = \frac{d}{3v}$.
When swimming upstream,the effective speed is $v_{up} = v_m - v_r = 2v - v = v$. The time taken is $t_2 = \frac{d}{v}$.
The total time taken is $t = t_1 + t_2 = \frac{d}{3v} + \frac{d}{v} = \frac{d + 3d}{3v} = \frac{4d}{3v}$.
Therefore,the ratio is $\frac{t}{t_0} = \frac{4d/3v}{d/v} = \frac{4}{3}$.

(A) Let $V$ be the speed of the bus and $V_o$ be the speed of the man. The distance between two consecutive buses is $d = V \cdot T$.
When the man moves in the direction of the bus,the relative speed is $V - V_o$. The time interval between buses passing him is $t_1 = \frac{d}{V - V_o} = \frac{VT}{V - V_o}$.
This implies $V - V_o = \frac{VT}{t_1} \quad \dots (1)$.
When the man moves in the opposite direction,the relative speed is $V + V_o$. The time interval between buses passing him is $t_2 = \frac{d}{V + V_o} = \frac{VT}{V + V_o}$.
This implies $V + V_o = \frac{VT}{t_2} \quad \dots (2)$.
Adding equations $(1)$ and $(2)$:
$(V - V_o) + (V + V_o) = \frac{VT}{t_1} + \frac{VT}{t_2}$
$2V = VT \left( \frac{1}{t_1} + \frac{1}{t_2} \right)$
$2 = T \left( \frac{t_1 + t_2}{t_1 t_2} \right)$
$T = \frac{2 t_1 t_2}{t_1 + t_2}$

(C) Let the total length of the two trains be $L$. Since they are moving in opposite directions,their relative velocity is $v_{rel} = v_1 + v_2$.
Given that they take $4 \ s$ to cross each other:
$L = (v_1 + v_2) \times 4$ --- $(i)$
When the speed of train $A$ is increased by $50 \%$,the new speed of train $A$ becomes $v_1' = v_1 + 0.5 v_1 = 1.5 v_1 = \frac{3}{2} v_1$.
The new relative velocity is $v_{rel}' = \frac{3}{2} v_1 + v_2$.
Given that they take $3 \ s$ to cross each other:
$L = (\frac{3}{2} v_1 + v_2) \times 3$ --- (ii)
Equating $(i)$ and (ii):
$4(v_1 + v_2) = 3(\frac{3}{2} v_1 + v_2)$
$4 v_1 + 4 v_2 = 4.5 v_1 + 3 v_2$
$v_2 = 0.5 v_1$
$\frac{v_1}{v_2} = \frac{1}{0.5} = \frac{2}{1}$.

(A) Let at time $t$,the position of $A$ be $\overrightarrow{r}_A = (0 + 0t)\hat{i} + (0 + 2t)\hat{j} = 2t\hat{j}$.
Let at time $t$,the position of $B$ be $\overrightarrow{r}_B = (0 + 2t)\hat{i} + (10 + 0t)\hat{j} = 2t\hat{i} + 10\hat{j}$.
The relative position vector $\overrightarrow{r}_{AB} = \overrightarrow{r}_B - \overrightarrow{r}_A = 2t\hat{i} + (10 - 2t)\hat{j}$.
The square of the distance $D^2 = |\overrightarrow{r}_{AB}|^2 = (2t)^2 + (10 - 2t)^2$.
$D^2 = 4t^2 + 100 + 4t^2 - 40t = 8t^2 - 40t + 100$.
For minimum distance,$\frac{d(D^2)}{dt} = 0$.
$16t - 40 = 0$.
$t = \frac{40}{16} = 2.5 \text{ s}$.
Since $\frac{d^2(D^2)}{dt^2} = 16 > 0$,the distance is minimum at $t = 2.5 \text{ s}$.

(A) The velocity of particle $A$ is constant along the $x$-axis: $\vec{V}_A = 1 \hat{i} \,m/s$.
For particle $B$ moving along the $y$-axis, the position is $y = ct^2$.
The velocity of particle $B$ is $\vec{V}_B = \frac{dy}{dt} \hat{j} = 2ct \hat{j}$.
At $t = 1 \,s$ and $c = 1 \,m/s^2$, the velocity of particle $B$ is $\vec{V}_B = 2(1)(1) \hat{j} = 2 \hat{j} \,m/s$.
The relative velocity of particle $A$ with respect to particle $B$ is $\vec{V}_{AB} = \vec{V}_A - \vec{V}_B = 1 \hat{i} - 2 \hat{j} \,m/s$.
The speed is the magnitude of the relative velocity: $|\vec{V}_{AB}| = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \,m/s$.

(A) Given:
Velocity of boy,$\overrightarrow{v_b} = 4 \ m \ s^{-1}$ and velocity of rain,$\overrightarrow{v_r} = 8 \ m \ s^{-1}$.
The velocity of rain with respect to the boy is $\overrightarrow{v_{rb}} = \overrightarrow{v_r} - \overrightarrow{v_b}$.
Let the vertical direction be the $y$-axis and the horizontal direction be the $x$-axis.
Rain velocity components: $v_{rx} = v_r \sin(45^{\circ}) = 8 \times \frac{1}{\sqrt{2}} = 4\sqrt{2} \ m \ s^{-1}$ and $v_{ry} = v_r \cos(45^{\circ}) = 8 \times \frac{1}{\sqrt{2}} = 4\sqrt{2} \ m \ s^{-1}$.
When the boy runs from west to east $(\overrightarrow{v_b} = 4 \hat{i})$:
$\overrightarrow{v_{rb}} = (v_{rx} \hat{i} - v_{ry} \hat{j}) - (4 \hat{i}) = (4\sqrt{2} - 4) \hat{i} - 4\sqrt{2} \hat{j}$.
$\tan \theta = \frac{|v_{rb,x}|}{|v_{rb,y}|} = \frac{4\sqrt{2} - 4}{4\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$.
When the boy runs from east to west $(\overrightarrow{v_b} = -4 \hat{i})$:
$\overrightarrow{v_{rb}} = (v_{rx} \hat{i} - v_{ry} \hat{j}) - (-4 \hat{i}) = (4\sqrt{2} + 4) \hat{i} - 4\sqrt{2} \hat{j}$.
$\tan \alpha = \frac{|v_{rb,x}|}{|v_{rb,y}|} = \frac{4\sqrt{2} + 4}{4\sqrt{2}} = \frac{\sqrt{2} + 1}{\sqrt{2}}$.
Therefore,the ratio $\frac{\tan \theta}{\tan \alpha} = \frac{(\sqrt{2} - 1)/\sqrt{2}}{(\sqrt{2} + 1)/\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$.
Rationalizing the denominator: $\frac{(\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{2 + 1 - 2\sqrt{2}}{2 - 1} = 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$.

(B) Let the velocity of the man be $\vec{v}_m = 6 \hat{i} \text{ km/h}$ and the velocity of the rain be $\vec{v}_r = -6\sqrt{3} \hat{j} \text{ km/h}$.
To protect himself,the man must hold the umbrella in the direction of the relative velocity of the rain with respect to the man,$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$.
$\vec{v}_{rm} = -6\sqrt{3} \hat{j} - 6 \hat{i} \text{ km/h}$.
The angle $\alpha$ with respect to the vertical is given by $\tan \alpha = \frac{|v_m|}{|v_r|} = \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Since $\tan \alpha = \frac{1}{\sqrt{3}}$,we have $\alpha = 30^{\circ}$.
Therefore,the man should hold his umbrella at an angle of $30^{\circ}$ with respect to the vertical.

(B) Let car $A$ be at the origin $(0, 0)$ moving north with velocity $\vec{V}_A = 25 \hat{j} \ km/hr$. Let car $B$ be at $(0, 50)$ moving west with velocity $\vec{V}_B = -25 \hat{i} \ km/hr$.
The relative velocity of $A$ with respect to $B$ is $\vec{V}_{AB} = \vec{V}_A - \vec{V}_B = 25 \hat{j} - (-25 \hat{i}) = 25 \hat{i} + 25 \hat{j} \ km/hr$.
The magnitude of relative velocity is $|\vec{V}_{AB}| = \sqrt{25^2 + 25^2} = 25\sqrt{2} \ km/hr$.
The relative position vector is $\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 0)\hat{i} + (0 - 50)\hat{j} = -50 \hat{j} \ km$.
The time taken to reach the closest approach is given by $t = -\frac{\vec{r}_{AB} \cdot \vec{V}_{AB}}{|\vec{V}_{AB}|^2}$.
$t = -\frac{(-50 \hat{j}) \cdot (25 \hat{i} + 25 \hat{j})}{(25\sqrt{2})^2} = -\frac{-1250}{1250} = 1 \ hr$.
Thus,the time taken is $60 \ min$.

(A) Given that,velocity of car $A$ is $\vec{v}_A = 30 \hat{i} \text{ km/h}$ (toward East).
For car $B$,velocity is $\vec{v}_B = 30 \hat{j} \text{ km/h}$ (toward North).
The velocity of car $B$ as measured from car $A$ is the relative velocity $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$.
$\vec{v}_{BA} = 30 \hat{j} - 30 \hat{i} \text{ km/h}$.
The magnitude of $\vec{v}_{BA}$ is $|\vec{v}_{BA}| = \sqrt{(-30)^2 + (30)^2} = \sqrt{900 + 900} = \sqrt{1800} = 30\sqrt{2} \approx 42 \text{ km/h}$.
The direction is given by $\tan \theta = \frac{v_{y}}{v_{x}} = \frac{30}{-30} = -1$. This corresponds to an angle of $135^{\circ}$ from the positive $x$-axis (East).
Since the vector points in the second quadrant (negative $x$,positive $y$),it is $45^{\circ}$ North of West.

(D) The man swims across the river by making strokes normal to the river current. The velocity of the man relative to the ground is the vector sum of his velocity in still water and the velocity of the river. However,the time taken to cross the river depends only on the component of his velocity perpendicular to the river flow.
Since the man swims normal to the current,his velocity component perpendicular to the river is $v_y = 4 \ km/h$.
The width of the river is $d = 1 \ km$.
The time taken to cross the river is given by:
$t = \frac{d}{v_y} = \frac{1 \ km}{4 \ km/h} = 0.25 \ h$
To convert this into minutes:
$t = 0.25 \times 60 \ min = 15 \ min$.

(C) The given situation is shown in the figure.
Velocity of car $A$ is along the negative $x$-axis:
$\vec{v}_A = -120 \hat{i} \text{ km/h}$
Velocity of car $B$ is along the negative $y$-axis:
$\vec{v}_B = -50 \hat{j} \text{ km/h}$
The relative velocity of car $B$ with respect to car $A$ is given by:
$\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$
$\vec{v}_{BA} = (-50 \hat{j}) - (-120 \hat{i})$
$\vec{v}_{BA} = 120 \hat{i} - 50 \hat{j}$
The relative speed is the magnitude of the relative velocity vector:
$|\vec{v}_{BA}| = \sqrt{(120)^2 + (-50)^2}$
$|\vec{v}_{BA}| = \sqrt{14400 + 2500}$
$|\vec{v}_{BA}| = \sqrt{16900}$
$|\vec{v}_{BA}| = 130 \text{ km/h}$

(C) The given situation is shown in the figure.
Velocity of river,$v_r = 3 \,m/s$.
Velocity of boat with respect to water,$v_b = 15 \,m/s$.
Width of river,$d = AB = 200 \,m$.
We need to calculate the horizontal drift,$BC$.
The time taken by the boat to cross the river is given by:
$t = \frac{d}{v_b} = \frac{200}{15} = \frac{40}{3} \,s$.
The horizontal distance covered by the boat due to the river flow is:
$BC = v_r \times t = 3 \times \frac{40}{3} = 40 \,m$.

(B) Let the velocity of rain be $\vec{v}_r$. The horizontal component of rain velocity is $v_{rx} = v_r \sin(30^{\circ})$ and the vertical component is $v_{ry} = v_r \cos(30^{\circ})$.
Given that the rain is falling at $30^{\circ}$ from the vertical with a speed of $40 \ m/s$,we have $v_r = 40 \ m/s$.
Thus,$v_{rx} = 40 \sin(30^{\circ}) = 40 \times 0.5 = 20 \ m/s$ and $v_{ry} = 40 \cos(30^{\circ}) = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \ m/s$.
The car is moving in the opposite direction to the horizontal component of the rain with a speed $v_c = 40 \ m/s$.
Let the velocity of the car be $\vec{v}_c = 40 \hat{i} \ m/s$. The velocity of rain relative to the car is $\vec{v}_{rc} = \vec{v}_r - \vec{v}_c$.
Assuming the rain's horizontal component is in the negative $x$-direction,$\vec{v}_r = -20 \hat{i} - 20\sqrt{3} \hat{j}$.
Then $\vec{v}_{rc} = (-20 \hat{i} - 20\sqrt{3} \hat{j}) - (40 \hat{i}) = -60 \hat{i} - 20\sqrt{3} \hat{j}$.
The angle $\alpha$ with the vertical is given by $\tan(\alpha) = \frac{|v_{rc,x}|}{|v_{rc,y}|} = \frac{60}{20\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Therefore,$\alpha = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.

(A) Given, speed of rain in vertical direction, $v_r = 30 \,m/s$.
Speed of the man, $v_m = 10 \,m/s$ (towards West).
The relative velocity of rain with respect to the man is $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$.
Since the man is moving towards the West, the relative velocity vector $\vec{v}_{rm}$ will be directed towards the West relative to the vertical.
Let $\theta$ be the angle with the vertical.
From the vector triangle, $\tan \theta = \frac{v_m}{v_r} = \frac{10}{30} = \frac{1}{3}$.
Therefore, $\theta = \tan^{-1}(1/3)$ towards the West direction.

(B) Given:
Width of the river $(w) = 200 \ m$
Velocity of the river $(v_r) = 2 \ m/s$
Velocity of the swimmer with respect to the river $(v_{sr}) = 1 \ m/s$
To cross the river in the shortest time,the swimmer should swim perpendicular to the river flow.
The time taken to cross the river is $t = \frac{w}{v_{sr}} = \frac{200 \ m}{1 \ m/s} = 200 \ s$.
During this time,the swimmer is carried downstream by the river flow.
The distance drifted downstream $(d)$ is given by $d = v_r \times t$.
$d = 2 \ m/s \times 200 \ s = 400 \ m$.
Therefore,the swimmer reaches the opposite bank at a distance of $400 \ m$ from the point directly opposite to the starting point.

$(\text{A})$ Let $\vec{v}_{r,g}$ be the velocity of rain with respect to the ground, $\vec{v}_{m,g}$ be the velocity of the man with respect to the ground, and $\vec{v}_{r,m}$ be the velocity of rain with respect to the man.
When the man is at rest, the rain falls at $30^{\circ}$ with the vertical. Thus, the horizontal component of $\vec{v}_{r,g}$ is $v_{r,g} \sin 30^{\circ}$ and the vertical component is $v_{r,g} \cos 30^{\circ}$.
When the man runs at $v_{m,g} = 10 \,km/h$, the rain appears to fall vertically. This means the horizontal component of the relative velocity $\vec{v}_{r,m} = \vec{v}_{r,g} - \vec{v}_{m,g}$ must be zero.
Therefore, the horizontal component of $\vec{v}_{r,g}$ must be equal to the velocity of the man:
$v_{r,g} \sin 30^{\circ} = v_{m,g}$
$v_{r,g} \times (1/2) = 10 \,km/h$
$v_{r,g} = 20 \,km/h$.

(B) The velocity of particle $A$ is given by $\vec{v}_A = 10 \hat{i} \ m/s$.
The velocity of particle $B$ can be resolved into components. Its magnitude is $20 \ m/s$ at an angle of $60^{\circ}$ with the $X$-axis:
$\vec{v}_B = (20 \cos 60^{\circ}) \hat{i} + (20 \sin 60^{\circ}) \hat{j}$
$\vec{v}_B = (20 \times 0.5) \hat{i} + (20 \times \frac{\sqrt{3}}{2}) \hat{j} = 10 \hat{i} + 10 \sqrt{3} \hat{j} \ m/s$.
The relative velocity of $B$ with respect to $A$ is $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$.
$\vec{v}_{BA} = (10 \hat{i} + 10 \sqrt{3} \hat{j}) - (10 \hat{i}) = 10 \sqrt{3} \hat{j} \ m/s$.
This result indicates a velocity of magnitude $10 \sqrt{3} \ m/s$ directed along the positive $Y$-axis,which is perpendicular to the $X$-axis.

(C) Given: Width of the river $d = 200 \ m$,velocity of river $v_R = 18 \ km/h = 18 \times \frac{5}{18} = 5 \ m/s$,velocity of boat in still water $v_{BR} = 36 \ km/h = 36 \times \frac{5}{18} = 10 \ m/s$.
To cross the river in minimum time,the boat must head perpendicular to the river flow.
The time taken to cross the river once is $t = \frac{d}{v_{BR}} = \frac{200}{10} = 20 \ s$.
For a round trip (going and coming back),the total time taken is $T = 20 + 20 = 40 \ s$.
During this time,the boat is carried downstream by the river flow. The velocity of the boat relative to the ground along the river bank is $v_R = 5 \ m/s$.
The displacement along the river bank is $x = v_R \times T = 5 \times 40 = 200 \ m$.

(C) Let the velocity of car $A$ be $v_A = 100 \text{ km/h}$ and car $B$ be $v_B = 80 \text{ km/h}$.
Relative velocity of $A$ with respect to $B$ is $v_{AB} = v_A - v_B = 100 - 80 = 20 \text{ km/h}$.
Converting this to $\text{m/s}$,we get $20 \times (5/18) = 5.55 \text{ m/s}$.
Let the velocity of the stone relative to car $B$ be $v$. Since the stone is thrown towards $A$,its velocity relative to the ground is $v_{sg} = v + v_B$.
The velocity of the stone relative to car $A$ is $v_{sa} = v_{sg} - v_A = (v + v_B) - v_A = v - (v_A - v_B) = v - 20 \text{ km/h}$.
Given that the speed of the stone relative to $A$ is $5 \text{ m/s}$,which is $5 \times (18/5) = 18 \text{ km/h}$.
Thus,$|v - 20| = 18$.
Since the stone must hit car $A$ (which is ahead),$v$ must be greater than $20 \text{ km/h}$.
Therefore,$v - 20 = 18$,which gives $v = 38 \text{ km/h}$.