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

(D) Let $\vec{V}_r$ be the velocity of rain and $\vec{V}_m$ be the velocity of the man.
When the man is running,the relative velocity of rain with respect to the man is $\vec{V}_{rm} = \vec{V}_r - \vec{V}_m$. Given that this is vertical,$\vec{V}_{rm} = -10\hat{j}$.
Thus,$\vec{V}_r - \vec{V}_m = -10\hat{j}$,which implies $\vec{V}_r = \vec{V}_m - 10\hat{j}$.
Since the man is running horizontally,let $\vec{V}_m = v\hat{i}$. So,$\vec{V}_r = v\hat{i} - 10\hat{j}$.
When the man stops,he observes the actual velocity of the rain $\vec{V}_r$,which makes an angle of $30^o$ with the vertical.
From the components of $\vec{V}_r$,$\tan 30^o = \frac{|v_x|}{|v_y|} = \frac{v}{10}$.
Therefore,$v = 10 \tan 30^o = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}}\,m/s$.

(B) For particle $A$ to catch particle $B$,the component of velocity of $A$ in the direction of $B$'s motion must be equal to the velocity of $B$.
Let the velocity of $A$ be $v$. The velocity of $B$ is $30 \ m/s$ in the vertical direction.
For $A$ to catch $B$,the vertical component of $A$'s velocity must equal $B$'s velocity:
$(v_y)_A = (v_y)_B$
$v \sin 37^{\circ} = 30 \ m/s$
Since $\sin 37^{\circ} = 3/5$,we have:
$v \times (3/5) = 30$
$v = 30 \times (5/3) = 50 \ m/s$
Now,the time taken to catch $B$ is the time required for $A$ to cover the horizontal distance of $200 \ m$ relative to $B$ at the relative horizontal velocity:
Time $t = \frac{\text{Distance}}{\text{Relative horizontal velocity}} = \frac{200}{v \cos 37^{\circ}}$
Since $\cos 37^{\circ} = 4/5$,we have:
$t = \frac{200}{50 \times (4/5)} = \frac{200}{40} = 5 \ s$.

(C) Let the velocity of the first ship be $\vec{v}_A = 10 \hat{i} \, km/hr$.
Let the velocity of the second ship be $\vec{v}_B = v_B \sin(30^o) \hat{i} + v_B \cos(30^o) \hat{j}$.
Since the second ship is always due north of the first ship,the relative velocity $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$ must have no component in the east-west direction (i.e.,the $\hat{i}$ component must be zero).
Therefore,$v_B \sin(30^o) - 10 = 0$.
$v_B \times (1/2) = 10$.
$v_B = 20 \, km/hr$.

(C) The relative velocity of particle $P_1$ with respect to $P_2$ is defined as $\vec{v}_{12} = \vec{v}_1 - \vec{v}_2$.
Using the triangle inequality for vectors,the magnitude of the relative velocity is given by $|\vec{v}_{12}| = |\vec{v}_1 + (-\vec{v}_2)|$.
According to the triangle inequality,$|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$.
Substituting $\vec{a} = \vec{v}_1$ and $\vec{b} = -\vec{v}_2$,we get $|\vec{v}_{12}| \leq |\vec{v}_1| + |-\vec{v}_2|$.
Since $|-\vec{v}_2| = |\vec{v}_2|$,this simplifies to $|\vec{v}_{12}| \leq v_1 + v_2$.
Therefore,the relative velocity $v_{12}$ cannot be greater than the sum of the magnitudes of the individual velocities $v_1 + v_2$.

(B) Let $v_m$ be the speed of the man in still water and $v_r$ be the speed of the river current.
Downstream speed: $v_m + v_r = \frac{10\,km}{2\,hrs} = 5\,km/h$ $...(i)$
Upstream speed: $v_m - v_r = \frac{30\,km}{10\,hrs} = 3\,km/h$ $...(ii)$
Adding equations $(i)$ and $(ii)$:
$2v_m = 8\,km/h \Rightarrow v_m = 4\,km/h$
Time taken to swim $40\,km$ in still water:
$t = \frac{Distance}{v_m} = \frac{40\,km}{4\,km/h} = 10\,hrs$.

(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 such that his resultant velocity is perpendicular to the river flow.
The effective velocity of the man across the river 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 the velocity of rain be $\vec{v}_r = -30\hat{j}\,ms^{-1}$ and the velocity of the bicycle be $\vec{v}_b = -12\hat{i}\,ms^{-1}$ (since she is moving from east to west).
The relative velocity of rain with respect to the woman is $\vec{v}_{rb} = \vec{v}_r - \vec{v}_b = -30\hat{j} - (-12\hat{i}) = 12\hat{i} - 30\hat{j}\,ms^{-1}$.
To protect herself from rain,she must hold the umbrella in the direction of the relative velocity of rain $\vec{v}_{rb}$.
The angle $\theta$ with the vertical is given by $\tan \theta = \frac{|v_b|}{|v_r|} = \frac{12}{30} = \frac{2}{5}$.
Thus,$\theta = \tan^{-1}(\frac{2}{5})$.
Since the relative velocity vector has a positive $x$-component (eastward relative to the woman) and a negative $y$-component (downward),the rain appears to come from the east. Therefore,she must tilt her umbrella towards the west to block the rain.

(A) To travel from $A$ to $B$ in a straight line while a wind of speed $v$ blows perpendicular to $AB$,the aeroplane must head at an angle such that its resultant velocity vector points directly along $AB$.
Let the resultant velocity of the aeroplane relative to the ground be $V_g$.
Using the Pythagorean theorem,$V^2 = V_g^2 + v^2$,which gives the ground speed $V_g = \sqrt{V^2 - v^2}$.
The time taken to travel from $A$ to $B$ is $t_1 = \frac{\ell}{V_g} = \frac{\ell}{\sqrt{V^2 - v^2}}$.
Similarly,the time taken to return from $B$ to $A$ is $t_2 = \frac{\ell}{V_g} = \frac{\ell}{\sqrt{V^2 - v^2}}$.
The total time for the round trip is $T = t_1 + t_2 = \frac{\ell}{\sqrt{V^2 - v^2}} + \frac{\ell}{\sqrt{V^2 - v^2}} = \frac{2\ell}{\sqrt{V^2 - v^2}}$.

(C) Let $\vec{V}_R$ be the actual velocity of the rain (vertically downward) and $\vec{V}_M$ be the velocity of the man ($12 \ km/h$ horizontally).
The relative velocity of the rain with respect to the man is $\vec{V}_{RM} = \vec{V}_R - \vec{V}_M$.
From the vector diagram,the angle $\theta = 60^{\circ}$ is between the vertical (direction of $\vec{V}_R$) and the relative velocity vector $\vec{V}_{RM}$.
Using trigonometry in the right-angled triangle formed by $\vec{V}_R$,$\vec{V}_M$,and $\vec{V}_{RM}$:
$\tan(60^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{V_M}{V_R}$
Substituting the known values:
$\sqrt{3} = \frac{12}{V_R}$
Solving for $V_R$:
$V_R = \frac{12}{\sqrt{3}} = \frac{12 \times \sqrt{3}}{3} = 4\sqrt{3} \ km/h$.

(C) Let the length of the escalator be $L$.
Speed of the person walking on the stalled escalator is $v_p = \frac{L}{60}$.
Speed of the escalator is $v_e = \frac{L}{40}$.
When the person walks on the moving escalator,their effective speed is $v_{eff} = v_p + v_e$.
$v_{eff} = \frac{L}{60} + \frac{L}{40} = L \left( \frac{2+3}{120} \right) = \frac{5L}{120} = \frac{L}{24}$.
The time taken to cover the distance $L$ is $t = \frac{L}{v_{eff}} = \frac{L}{L/24} = 24\,s$.

(A) The total distance to be covered by the passenger train to completely cross the freight train is the sum of their lengths: $D = 60\, m + 120\, m = 180\, m = 0.18\, km$.
$(i)$ When moving in the same direction,the relative speed is $v_{rel} = 80 - 30 = 50\, km/hr$.
The time taken is $t_1 = \frac{D}{v_{rel}} = \frac{0.18}{50}\, hr$.
$(ii)$ When moving in the opposite direction,the relative speed is $v_{rel} = 80 + 30 = 110\, km/hr$.
The time taken is $t_2 = \frac{D}{v_{rel}} = \frac{0.18}{110}\, hr$.
The ratio of the times is $\frac{t_1}{t_2} = \frac{0.18 / 50}{0.18 / 110} = \frac{110}{50} = \frac{11}{5} = 2.2$.

(C) Let the position of ship $A$ be at the origin $(0,0)$.
The initial position vectors are $\vec{r}_A = 0\hat{i} + 0\hat{j}$ and $\vec{r}_B = 80\hat{i} + 150\hat{j}$.
The velocity vectors are $\vec{v}_A = 30\hat{i} + 50\hat{j}$ and $\vec{v}_B = -10\hat{i}$.
The relative position vector is $\vec{r}_{BA} = \vec{r}_B - \vec{r}_A = 80\hat{i} + 150\hat{j}$.
The relative velocity vector is $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = (-10\hat{i}) - (30\hat{i} + 50\hat{j}) = -40\hat{i} - 50\hat{j}$.
The time $t$ at which the distance is minimum is given by $t = -\frac{\vec{r}_{BA} \cdot \vec{v}_{BA}}{|\vec{v}_{BA}|^2}$.
Substituting the values:
$t = -\frac{(80\hat{i} + 150\hat{j}) \cdot (-40\hat{i} - 50\hat{j})}{(-40)^2 + (-50)^2}$
$t = -\frac{(80 \times -40) + (150 \times -50)}{1600 + 2500}$
$t = -\frac{-3200 - 7500}{4100} = \frac{10700}{4100} = \frac{107}{41} \approx 2.61\,\text{hrs}$.
Rounding to the nearest given option,the time is $2.6\,\text{hrs}$.

(C) To cross the river straight,the component of the swimmer's velocity perpendicular to the river flow must cancel out the river's velocity component in the opposite direction.
Let $\theta$ be the angle the swimmer makes with the line perpendicular to the river flow.
The velocity of the river is $v_r = 2\,km/h$.
The velocity of the swimmer is $v_s = 4\,km/h$.
For the swimmer to move straight across,the horizontal component of the swimmer's velocity must equal the river's velocity:
$v_s \sin \theta = v_r$
$4 \sin \theta = 2$
$\sin \theta = \frac{2}{4} = \frac{1}{2}$
$\theta = 30^\circ$ (with respect to the perpendicular line).
The angle with respect to the direction of the river flow is $90^\circ + \theta = 90^\circ + 30^\circ = 120^\circ$.

(D) Let the velocity of the belt with respect to the ground be $v_{BG} = 4 \, km \, h^{-1}$.
Let the velocity of the child with respect to the belt be $v_{CB} = 9 \, km \, h^{-1}$.
We need to find the velocity of the child with respect to the ground $(v_{CG})$ when the child is running in the direction of the belt's motion.
According to the relative velocity formula: $v_{CG} = v_{CB} + v_{BG}$.
Substituting the given values:
$v_{CG} = 9 \, km \, h^{-1} + 4 \, km \, h^{-1} = 13 \, km \, h^{-1}$.
Thus,for an observer on a stationary platform,the speed of the child running in the direction of motion of the belt is $13 \, km \, h^{-1}$.

(C) Let the speed of the child with respect to the belt be $v_{cb} = 9\,km/h$.
Let the speed of the belt with respect to the ground (platform) be $v_{bg} = 4\,km/h$.
We need to find the speed of the child with respect to the ground $(v_{cg})$ when the child is running in the direction of the belt's motion.
Using the concept of relative velocity: $v_{cg} = v_{cb} + v_{bg}$.
Substituting the given values: $v_{cg} = 9\,km/h + 4\,km/h = 13\,km/h$.
Therefore,for an observer on the platform,the speed of the child in the direction of motion of the belt is $13\,km/h$.

(A) Let the speed of cars $A$ and $B$ be $v_A = v_B = 45\,km/h$ in the same direction.
Let the speed of car $C$ be $v_C = 36\,km/h$ in the opposite direction.
The relative speed of car $C$ with respect to cars $A$ and $B$ is $v_{rel} = v_A + v_C = 45 + 36 = 81\,km/h$.
Since car $C$ meets car $A$ and car $B$ with a time interval of $\Delta t = 5\,min = \frac{5}{60}\,h = \frac{1}{12}\,h$,the distance $d$ between car $A$ and car $B$ is given by the product of their relative speed and the time interval.
$d = v_{rel} \times \Delta t = 81 \times \frac{1}{12} = 6.75\,km$.

(D) Let the positive direction of the $x$-axis be from south to north.
Velocity of the train,$v_{T} = +10 \; m/s$.
Velocity of the parrot,$v_{P} = -5 \; m/s$.
The relative velocity of the parrot with respect to the train is given by $v_{PT} = v_{P} - v_{T}$.
$v_{PT} = (-5 \; m/s) - (+10 \; m/s) = -15 \; m/s$.
The magnitude of the relative velocity is $15 \; m/s$,which means the parrot appears to move at $15 \; m/s$ from north to south relative to the train.
The length of the train to be covered is $L = 150 \; m$.
Therefore,the time taken by the parrot to cross the train is $t = \frac{L}{|v_{PT}|} = \frac{150 \; m}{15 \; m/s} = 10 \; s$.

(A) Let the velocity of the man be $\vec{v}_m = 5 \, m/s$ (horizontal) and the velocity of the rain be $\vec{v}_r = v \, m/s$ (vertically downwards).
The velocity of rain relative to the man is $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$.
This relative velocity vector $\vec{v}_{rm}$ makes an angle of $45^{\circ}$ with the vertical.
Using the vector components,we have $\tan 45^{\circ} = \frac{|v_m|}{|v_r|} = \frac{5}{v}$.
Since $\tan 45^{\circ} = 1$,we get $1 = \frac{5}{v}$,which implies $v = 5 \, m/s$.

(A) Given:
Velocity of particle $A$,$\vec{v}_A = 3\hat{j}\,km/h$
Velocity of particle $B$,$\vec{v}_B = -4\hat{i}\,km/h$
The relative velocity of $A$ with respect to $B$ is given by $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.
$\vec{v}_{AB} = 3\hat{j} - (-4\hat{i}) = 4\hat{i} + 3\hat{j}\,km/h$.
The magnitude of relative velocity is $|\vec{v}_{AB}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\,km/h$.
To find the direction,let $\alpha$ be the angle with the east direction (positive $x$-axis).
$\tan \alpha = \frac{v_{AB,y}}{v_{AB,x}} = \frac{3}{4}$.
Since $\tan 37^{\circ} = 3/4$,we have $\alpha = 37^{\circ}$.
Thus,the direction is $37^{\circ}$ north of east.
Therefore,the correct option is $A$.

(A) To cross a river in the shortest time,the swimmer must maximize their velocity component perpendicular to the river flow.
Let the velocity of the man in still water be $v_m = 20\,m/min$ and the velocity of the river be $v_r = 8\,m/min$.
The time taken to cross the river of width $d$ is given by $t = \frac{d}{v_m \cos \theta}$,where $\theta$ is the angle with the normal to the river bank.
To minimize $t$,the denominator $v_m \cos \theta$ must be maximized.
This occurs when $\cos \theta = 1$,which means $\theta = 0^{\circ}$.
Therefore,the man should swim perpendicular to the river flow,which is due north.

(C) Let the direction of the thief's car be positive.
Velocity of the thief's car with respect to the ground: $v_{TG} = 10\,m/s$.
Velocity of the police van with respect to the ground: $v_{PG} = 5\,m/s$.
Muzzle velocity of the bullet with respect to the police van: $v_{BP} = 72\,km/h = 72 \times \frac{5}{18} = 20\,m/s$.
Velocity of the bullet with respect to the ground: $v_{BG} = v_{BP} + v_{PG} = 20 + 5 = 25\,m/s$.
Velocity of the bullet with respect to the thief's car: $v_{BT} = v_{BG} - v_{TG} = 25 - 10 = 15\,m/s$.
Thus,the bullet hits the car with a speed of $15\,m/s$.

(B) Let the speeds of the two cars be $v_1$ and $v_2$ (where $v_1 > v_2$).
When travelling in opposite directions,the relative speed is the sum of their individual speeds:
$v_1 + v_2 = \frac{8\,m}{1\,s} = 8\,ms^{-1}$ (Equation $1$)
When travelling in the same direction,the relative speed is the difference of their individual speeds:
$v_1 - v_2 = \frac{0.8\,m}{1\,s} = 0.8\,ms^{-1}$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(v_1 + v_2) + (v_1 - v_2) = 8 + 0.8$
$2v_1 = 8.8$
$v_1 = 4.4\,ms^{-1}$
Substituting $v_1$ in Equation $1$:
$4.4 + v_2 = 8$
$v_2 = 8 - 4.4 = 3.6\,ms^{-1}$
Thus,the speeds of the two cars are $4.4\,ms^{-1}$ and $3.6\,ms^{-1}$.

(A) Let the speed of each train be $v$.
Since the trains are moving in opposite directions,let the velocity of the first train be $\vec{v}_1 = v \hat{i}$ and the second train be $\vec{v}_2 = -v \hat{i}$.
The wind is blowing with speed $u$ along the track,so its velocity is $\vec{v}_w = u \hat{i}$.
The relative velocity of the first train with respect to the wind is $\vec{v}_{1w} = \vec{v}_1 - \vec{v}_w = (v - u) \hat{i}$.
The relative velocity of the second train with respect to the wind is $\vec{v}_{2w} = \vec{v}_2 - \vec{v}_w = (-v - u) \hat{i} = -(v + u) \hat{i}$.
The ratio of the magnitudes of these relative velocities is given as $1:2$.
Therefore,$\frac{|v - u|}{|-(v + u)|} = \frac{1}{2}$.
Assuming $v > u$,we have $\frac{v - u}{v + u} = \frac{1}{2}$.
Cross-multiplying gives $2(v - u) = v + u$,which simplifies to $2v - 2u = v + u$.
Solving for $v$,we get $v = 3u$.

(A) To ensure the water drops enter the tube without touching its sides,the tube must be aligned with the direction of the relative velocity of the rain with respect to the boy.
Let the velocity of the boy be $\vec{v}_b = v \hat{i}$ and the velocity of the rain be $\vec{v}_r = -u \hat{j}$.
The relative velocity of the rain with respect to the boy is $\vec{v}_{rb} = \vec{v}_r - \vec{v}_b = -u \hat{j} - v \hat{i}$.
The angle $\theta$ that the resultant velocity vector makes with the vertical is given by $\tan \theta = \frac{|v_x|}{|v_y|} = \frac{v}{u}$.
Therefore,the angle with the vertical is $\theta = \tan ^{-1}\left(\frac{v}{u}\right)$.

(D) Let the length of the escalator be $s$.
The speed of the person with respect to the escalator is $v_p = \frac{s}{90}$.
The speed of the escalator is $v_e = \frac{s}{60}$.
When the person walks up the moving escalator,his effective speed is $v_{eff} = v_p + v_e$.
$v_{eff} = \frac{s}{90} + \frac{s}{60} = s \left( \frac{2 + 3}{180} \right) = \frac{5s}{180} = \frac{s}{36}$.
The time taken to cover the distance $s$ is $t = \frac{s}{v_{eff}} = \frac{s}{s/36} = 36 \, s$.

(B) Let the velocity of ship $X$ be $\vec{v}_X = v\hat{j}$.
Given that the velocity of $Y$ relative to $X$ is $\vec{v}_{YX} = -v\hat{i}$ (since it appears to move west).
The relative velocity is defined as $\vec{v}_{YX} = \vec{v}_Y - \vec{v}_X$.
Therefore,the actual velocity of $Y$ is $\vec{v}_Y = \vec{v}_{YX} + \vec{v}_X$.
Substituting the values: $\vec{v}_Y = -v\hat{i} + v\hat{j}$.
The magnitude is $|\vec{v}_Y| = \sqrt{(-v)^2 + v^2} = \sqrt{2}v$.
The direction is north-west because the vector has a negative $x$-component (west) and a positive $y$-component (north).

(C) Let $v_r$ be the velocity of the swimmer with respect to the river.
The swimmer heads at an angle of $120^{\circ}$ with the river flow $v$. The component of $v_r$ perpendicular to the river flow is $v_r \sin(120^{\circ}-90^{\circ}) = v_r \sin 30^{\circ} = v_r/2$ is incorrect based on the geometry.
Actually,the angle with the normal to the river flow is $120^{\circ}-90^{\circ} = 30^{\circ}$.
The time taken to cross the river is $t = \frac{d}{v_r \cos 30^{\circ}} = \frac{d}{v_r (\sqrt{3}/2)} = \frac{2d}{\sqrt{3} v_r}$.
The drift in the direction of the river flow is given by $x = (v - v_r \sin 30^{\circ})t$.
Given $x = d/2$,we have $d/2 = (v - v_r/2) \left( \frac{2d}{\sqrt{3} v_r} \right)$.
Simplifying,$1/2 = \frac{2v - v_r}{\sqrt{3} v_r} \implies \sqrt{3} v_r = 4v - 2v_r$.
$v_r(2 + \sqrt{3}) = 4v \implies v_r = \frac{4v}{2 + \sqrt{3}}$.
Rationalizing the denominator: $v_r = \frac{4v(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{4v(2 - \sqrt{3})}{4 - 3} = 4(2 - \sqrt{3})v$.

(D) Let the velocity of rain be $\vec{v}_r = -v \hat{j}$,where $v$ is the speed of rain.
Let the velocity of the train be $\vec{v}_t = v \hat{i}$ (moving towards the east).
The velocity of rain relative to the train is $\vec{v}_{rt} = \vec{v}_r - \vec{v}_t$.
Substituting the values,we get $\vec{v}_{rt} = -v \hat{j} - v \hat{i} = -v(\hat{i} + \hat{j})$.
The direction of this vector is in the third quadrant,which corresponds to the direction between south and west. However,looking at the relative motion from the perspective of a passenger,the rain appears to come from the opposite direction of the resultant velocity vector.
Since the horizontal component is towards the west and the vertical component is downwards,the rain appears to fall at an angle of $45^o$ to the vertical towards the west.

(B) Let the distance across the lake be $S$.
On a calm day,the speed of the boat is $V$. The time taken to go and return is $T_0 = \frac{S}{V} + \frac{S}{V} = \frac{2S}{V}$.
On a rough day,the current speed is $v$. The speed during the onward journey is $(V + v)$ and during the return journey is $(V - v)$.
The total time taken is $T = \frac{S}{V + v} + \frac{S}{V - v}$.
$T = S \left( \frac{V - v + V + v}{V^2 - v^2} \right) = \frac{2SV}{V^2 - v^2}$.
Now,calculating the ratio $T / T_0$:
$T / T_0 = \frac{2SV / (V^2 - v^2)}{2S / V} = \frac{V^2}{V^2 - v^2}$.
Dividing the numerator and denominator by $V^2$,we get $T / T_0 = \frac{1}{1 - v^2 / V^2}$.

(C) Let $L$ be the length of the lake and $V$ be the velocity of the boat relative to the water.
On a quiet day,the velocity of the water is $0$. The time taken for the round trip is:
$t_{Q} = \frac{L}{V} + \frac{L}{V} = \frac{2L}{V} \quad ...(i)$
On a rough day,let $v$ be the velocity of the uniform current. During the onward journey,the effective speed is $(V + v)$,and during the return journey,the effective speed is $(V - v)$.
The time taken for the round trip on a rough day is:
$t_{R} = \frac{L}{V + v} + \frac{L}{V - v} = \frac{L(V - v) + L(V + v)}{V^{2} - v^{2}} = \frac{2LV}{V^{2} - v^{2}} = \frac{2L}{V(1 - (v/V)^{2})} \quad ...(ii)$
Comparing equations $(i)$ and $(ii)$:
$t_{R} = t_{Q} \times \frac{1}{1 - (v/V)^{2}}$
Since $(v/V)^{2} > 0$,the denominator $1 - (v/V)^{2} < 1$,which implies that $\frac{1}{1 - (v/V)^{2}} > 1$.
Therefore,$t_{R} > t_{Q}$. The time required on a rough day is more.

(A) Given,speed of the first car $v_1 = 18\,km/h$.
Speed of the second car $v_2 = 27\,km/h$.
Since the cars are moving towards each other,their relative speed is $v_{rel} = v_1 + v_2 = 18 + 27 = 45\,km/h$.
The initial distance between the cars is $d = 36\,km$.
The time taken for the cars to meet is $t = \frac{d}{v_{rel}} = \frac{36}{45} = 0.8\,h$.
The bird flies continuously at a constant speed $v_b = 36\,km/h$ until the cars meet.
Therefore,the total distance covered by the bird is $D = v_b \times t = 36 \times 0.8 = 28.8\,km$.

(A) The time taken to cross a river of width $d$ is given by $t = \frac{d}{v_y}$,where $v_y$ is the component of the boat's velocity relative to the river in the direction normal to the river flow.
If the magnitudes of the velocities of two boats relative to the river are the same $(v_1 = v_2 = v)$,and they are directed at the same angle $\theta$ with respect to the normal to the river bank,then their normal components are $v_y = v \cos \theta$.
Since $v$ and $\theta$ are the same for both boats,their normal components $v_y$ are identical.
Therefore,both boats will cross the river in the same time $t = \frac{d}{v \cos \theta}$,even if they follow different paths due to the river's flow.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) Let the speed of the swimmer in still water be $v = 20 \; m/s$ and the speed of the river be $u = 10 \; m/s$.
To cross the river along the shortest path (perpendicular to the river flow),the swimmer must swim at an angle $\theta$ with respect to the north (perpendicular to the bank) such that the horizontal component of his velocity cancels the river's velocity.
From the geometry of the velocity vector triangle,we have:
$\sin \theta = \frac{u}{v}$
Substituting the given values:
$\sin \theta = \frac{10}{20} = \frac{1}{2}$
Therefore,$\theta = \arcsin(1/2) = 30^{\circ}$.
Thus,the swimmer should make his strokes at an angle of $30^{\circ}$ west of north.

(C) The velocity of the first particle is given by the derivative of its position with respect to time: $v_x = \frac{dx}{dt} = 8 - 6t$. At $t = 1 \; s$,$v_x = 8 - 6(1) = 2 \; m/s$. Thus,$\vec{v}_1 = 2 \hat{i} \; m/s$.
The velocity of the second particle is given by the derivative of its position with respect to time: $v_y = \frac{dy}{dt} = -24t^2$. At $t = 1 \; s$,$v_y = -24(1)^2 = -24 \; m/s$. Thus,$\vec{v}_2 = -24 \hat{j} \; m/s$.
The velocity of the second particle relative to the first particle is $\vec{v}_{21} = \vec{v}_2 - \vec{v}_1 = -2 \hat{i} - 24 \hat{j}$.
The speed of the second particle in the frame of the first is the magnitude of the relative velocity: $|\vec{v}_{21}| = \sqrt{(-2)^2 + (-24)^2} = \sqrt{4 + 576} = \sqrt{580}$.
Given that the speed is $\sqrt{v}$,we have $\sqrt{v} = \sqrt{580}$,which implies $v = 580$.

(N/A) The line of sight is defined as an imaginary line joining an object and an observer's eye.
When we observe nearby stationary objects such as trees or houses while sitting in a moving train,the angle subtended by these objects at our eye changes very rapidly as we move past them. This rapid change in the direction of the line of sight makes them appear to move quickly in the opposite direction.
On the other hand,for distant objects like hilltops,the Moon,or stars,the distance is so large that the change in the angle of the line of sight is negligible even as the train covers a significant distance. Consequently,these objects appear to be stationary or moving slowly with the observer.

(A) The relation $\tan \theta = v$ is incorrect.
Dimensional Analysis:
Dimension of $L.H.S = \tan \theta = [M^0 L^0 T^0]$ (dimensionless).
Dimension of $R.H.S = v = [L T^{-1}]$.
Since the dimensions of $L.H.S$ and $R.H.S$ do not match,the equation is dimensionally inconsistent.
To make it dimensionally correct,the $R.H.S$ must also be dimensionless. This can be achieved by dividing $v$ by the speed of the rain $v'$,which is also a velocity.
Therefore,the correct relation is $\tan \theta = \frac{v}{v'}$,where $v'$ is the speed of the rain.

(D) Let the positive direction of the $x-$axis be from south to north.
Then,the velocity of train $A$ is $v_{A} = +54 \; km \; h^{-1} = 54 \times \frac{5}{18} \; m \; s^{-1} = +15 \; m \; s^{-1}.$
The velocity of train $B$ is $v_{B} = -90 \; km \; h^{-1} = -90 \times \frac{5}{18} \; m \; s^{-1} = -25 \; m \; s^{-1}.$
The relative velocity of $B$ with respect to $A$ is given by $v_{BA} = v_{B} - v_{A}.$
Substituting the values,$v_{BA} = -25 \; m \; s^{-1} - (+15 \; m \; s^{-1}) = -40 \; m \; s^{-1}.$
The negative sign indicates that the train $B$ appears to $A$ to move with a speed of $40 \; m \; s^{-1}$ from north to south.

(A) Let the positive direction of the $x$-axis be from south to north.
The velocity of train $A$ is $v_{A} = +54 \; km \; h^{-1} = 54 \times \frac{5}{18} \; m \; s^{-1} = 15 \; m \; s^{-1}$.
The velocity of train $B$ is $v_{B} = -90 \; km \; h^{-1} = -90 \times \frac{5}{18} \; m \; s^{-1} = -25 \; m \; s^{-1}$.
The velocity of the ground with respect to train $B$ is given by $v_{GB} = v_{G} - v_{B}$.
Since the ground is stationary,$v_{G} = 0 \; m \; s^{-1}$.
Therefore,$v_{GB} = 0 - (-25 \; m \; s^{-1}) = +25 \; m \; s^{-1}$.
Thus,the velocity of the ground with respect to train $B$ is $25 \; m \; s^{-1}$ towards the north.

(B) Let the positive direction of the $x$-axis be from south to north.
The velocity of train $A$ is $v_{A} = +54\; km\; h^{-1} = 54 \times \frac{5}{18}\; m\; s^{-1} = +15\; m\; s^{-1}$.
The velocity of the monkey with respect to train $A$ is $v_{MA} = -18\; km\; h^{-1}$ (since it moves against the motion of the train).
Converting this to $m\; s^{-1}$: $v_{MA} = -18 \times \frac{5}{18}\; m\; s^{-1} = -5\; m\; s^{-1}$.
We know that the relative velocity of the monkey with respect to train $A$ is given by $v_{MA} = v_{M} - v_{A}$,where $v_{M}$ is the velocity of the monkey with respect to the ground.
Substituting the values: $-5\; m\; s^{-1} = v_{M} - 15\; m\; s^{-1}$.
Therefore,$v_{M} = 15\; m\; s^{-1} - 5\; m\; s^{-1} = 10\; m\; s^{-1}$.

(C) Let the velocity of the jet airplane with respect to the ground be $v_p = 500 \; km \; h^{-1}$.
Let the velocity of the combustion products with respect to the jet airplane be $v_{cp/p} = -1500 \; km \; h^{-1}$ (since they are ejected in the opposite direction to the plane's motion).
The velocity of the combustion products with respect to the ground $v_{cp}$ is given by the relative velocity formula:
$v_{cp/p} = v_{cp} - v_p$
Substituting the values:
$-1500 = v_{cp} - 500$
$v_{cp} = -1500 + 500 = -1000 \; km \; h^{-1}$.
The speed is the magnitude of the velocity,which is $1000 \; km \; h^{-1}$.

(B) First,convert all speeds to $m/s$:
$v_A = 36 \times (5/18) = 10 \; m/s$.
$v_B = 54 \times (5/18) = 15 \; m/s$.
$v_C = 54 \times (5/18) = 15 \; m/s$.
Relative velocity of $B$ w.r.t $A$ is $v_{BA} = v_B - v_A = 15 - 10 = 5 \; m/s$.
Relative velocity of $C$ w.r.t $A$ is $v_{CA} = v_C - (-v_A) = 15 + 10 = 25 \; m/s$.
Distance $s = 1 \; km = 1000 \; m$.
Time taken by car $C$ to reach car $A$ is $t = s / v_{CA} = 1000 / 25 = 40 \; s$.
To avoid an accident,car $B$ must cover $1000 \; m$ in $t = 40 \; s$.
Using the equation of motion $s = u_{BA}t + (1/2)at^2$:
$1000 = 5 \times 40 + (1/2) \times a \times (40)^2$.
$1000 = 200 + 800a$.
$800 = 800a$.
$a = 1 \; m/s^2$.

(C) Let $V$ be the speed of the bus. The speed of the cyclist is $v = 20 \; km/h$.
When the bus moves in the direction of the cyclist,the relative speed is $(V - 20) \; km/h$. The time interval between buses passing the cyclist is $18 \; min = 18/60 \; h$. The distance between two consecutive buses is $V \times (T/60)$. Thus,$(V - 20) \times (18/60) = V \times (T/60) \implies (V - 20) \times 18 = VT \quad \dots (i)$.
When the bus moves in the opposite direction,the relative speed is $(V + 20) \; km/h$. The time interval between buses passing the cyclist is $6 \; min = 6/60 \; h$. Thus,$(V + 20) \times (6/60) = V \times (T/60) \implies (V + 20) \times 6 = VT \quad \dots (ii)$.
Equating $(i)$ and $(ii)$:
$(V - 20) \times 18 = (V + 20) \times 6$
$3(V - 20) = V + 20$
$3V - 60 = V + 20$
$2V = 80 \implies V = 40 \; km/h$.
Substituting $V = 40$ in $(ii)$:
$(40 + 20) \times 6 = 40 \times T$
$60 \times 6 = 40T$
$360 = 40T \implies T = 9 \; min$.

(C) First,convert the speeds from $km/h$ to $m/s$ by multiplying by $\frac{5}{18}$.
Speed of the police van,$v_p = 30 \times \frac{5}{18} = 8.33\; m/s$.
Speed of the thief's car,$v_t = 192 \times \frac{5}{18} = 53.33\; m/s$.
The muzzle speed of the bullet relative to the police van is $v_{b/p} = 150\; m/s$.
The velocity of the bullet with respect to the ground is $v_b = v_{b/p} + v_p = 150 + 8.33 = 158.33\; m/s$.
The relative velocity of the bullet with respect to the thief's car is $v_{b/t} = v_b - v_t$.
$v_{b/t} = 158.33 - 53.33 = 105\; m/s$.

(N/A) Speed of the belt,$v_{B} = 4 \; km \; h^{-1}$.
Speed of the child with respect to the belt,$v_{c} = 9 \; km \; h^{-1}$.
$(a)$ Since the child is running in the same direction as the motion of the belt,his speed as observed by the stationary observer is:
$v_{a} = v_{c} + v_{B} = 9 + 4 = 13 \; km \; h^{-1}$.
$(b)$ Since the child is running opposite to the direction of motion of the belt,his speed as observed by the stationary observer is:
$v_{b} = v_{c} - v_{B} = 9 - 4 = 5 \; km \; h^{-1}$.
$(c)$ Distance between the parents,$d = 50 \; m = 0.05 \; km$.
Since both parents are on the moving belt,the relative speed of the child with respect to the parents is always $9 \; km \; h^{-1}$.
Converting speed to $m \; s^{-1}$: $9 \; km \; h^{-1} = 9 \times \frac{5}{18} = 2.5 \; m \; s^{-1}$.
Time taken,$t = \frac{d}{v_{c}} = \frac{50 \; m}{2.5 \; m \; s^{-1}} = 20 \; s$.
If the motion is viewed by one of the parents,the answers for $(a)$ and $(b)$ will alter because the parents are in the same frame of reference as the belt. For the parents,the speed of the child is always $9 \; km \; h^{-1}$ regardless of the direction. The time taken remains unaltered.

(N/A) The velocity of the rain $(v_r)$ and the wind $(v_w)$ are represented by vectors as shown in the figure. The rain falls vertically downwards,and the wind blows from east to west.
Using the rule of vector addition,the resultant velocity $R$ of the rain relative to the ground is the vector sum of $v_r$ and $v_w$.
The magnitude of the resultant velocity $R$ is given by:
$R = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2} \; m s^{-1} = \sqrt{1225 + 144} \; m s^{-1} = \sqrt{1369} \; m s^{-1} = 37 \; m s^{-1}$.
The direction $\theta$ that the resultant velocity $R$ makes with the vertical is given by:
$\tan \theta = \frac{v_w}{v_r} = \frac{12}{35} \approx 0.343$.
$\theta = \tan^{-1}(0.343) \approx 19^{\circ}$.
Since the wind is blowing towards the west,the rain appears to come from the west. Therefore,the boy should hold his umbrella in the vertical plane at an angle of about $19^{\circ}$ with the vertical towards the west.

(N/A) Let $v_b$ be the velocity of the motorboat (directed North) and $v_c$ be the velocity of the water current (directed $60^{\circ}$ East of South). The angle between $v_b$ and $v_c$ is $\theta = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Using the law of cosines,the magnitude of the resultant velocity $R$ is:
$R = \sqrt{v_b^2 + v_c^2 + 2v_b v_c \cos(120^{\circ})}$
$R = \sqrt{25^2 + 10^2 + 2(25)(10)(-0.5)}$
$R = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 21.8\; km/h$.
To find the direction $\phi$ of the resultant velocity with respect to the North direction,we use the law of sines:
$\frac{R}{\sin(120^{\circ})} = \frac{v_c}{\sin(\phi)}$
$\sin(\phi) = \frac{v_c \sin(120^{\circ})}{R} = \frac{10 \times 0.866}{21.8} \approx 0.397$
$\phi = \arcsin(0.397) \approx 23.4^{\circ}$.
Thus,the resultant velocity is approximately $21.8\; km/h$ at an angle of $23.4^{\circ}$ East of North.

(N/A) Let $v_r$ represent the velocity of rain and $v_b$ represent the velocity of the bicycle,both with respect to the ground.
Since the woman is riding a bicycle,the velocity of rain as experienced by her is the velocity of rain relative to the bicycle,given by:
$v_{rb} = v_r - v_b$
This relative velocity vector $v_{rb}$ makes an angle $\theta$ with the vertical,as shown in the figure.
It is given by:
$\tan \theta = \frac{|v_b|}{|v_r|} = \frac{12}{35} \approx 0.343$
$\theta = \tan^{-1}(0.343) \approx 19^{\circ}$
Therefore,the woman should hold her umbrella at an angle of about $19^{\circ}$ with the vertical towards the west.

(B) Let $v_r$ be the velocity of the rain falling vertically downwards,$v_r = 30\; m/s$.
Let $v_c$ be the velocity of the cyclist moving from north to south,$v_c = 10\; m/s$.
The relative velocity of the rain with respect to the woman is given by $\vec{v}_{rw} = \vec{v}_r - \vec{v}_c$.
To protect herself from the rain,the woman must hold her umbrella in the direction of the relative velocity $\vec{v}_{rw}$.
The angle $\theta$ that the relative velocity vector makes with the vertical is given by:
$\tan \theta = \frac{|\vec{v}_c|}{|\vec{v}_r|} = \frac{10}{30} = \frac{1}{3}$.
Therefore,$\theta = \tan^{-1}\left(\frac{1}{3}\right)$.
The woman should hold her umbrella at an angle of $\tan^{-1}\left(\frac{1}{3}\right)$ with the vertical towards the south.

(A) Speed of the man in still water,$v_m = 4.0 \; km/h$.
Width of the river,$d = 1.0 \; km$.
Since the man swims normal to the river current,his velocity component across the river is $v_m = 4.0 \; km/h$.
Time taken to cross the river,$t = \frac{d}{v_m} = \frac{1.0 \; km}{4.0 \; km/h} = 0.25 \; h$.
Converting to minutes,$t = 0.25 \times 60 = 15 \; min$.
Speed of the river,$v_r = 3.0 \; km/h$.
The distance covered down the river (drift) is $x = v_r \times t = 3.0 \; km/h \times 0.25 \; h = 0.75 \; km$.
Converting to meters,$x = 0.75 \times 1000 = 750 \; m$.

(D) Velocity of the boat,$v_{b} = 51 \; km/h$ (towards North).
Velocity of the wind,$v_{w} = 72 \; km/h$ (towards North-East).
The flag flutters in the direction of the relative velocity of the wind with respect to the boat,given by $\vec{v}_{wb} = \vec{v}_{w} - \vec{v}_{b} = \vec{v}_{w} + (-\vec{v}_{b})$.
Here,$\vec{v}_{w}$ makes an angle of $45^{\circ}$ with the North (or East) direction. The vector $(-\vec{v}_{b})$ is directed towards the South.
The angle between $\vec{v}_{w}$ and $(-\vec{v}_{b})$ is $90^{\circ} + 45^{\circ} = 135^{\circ}$.
Using the formula for the direction $\beta$ of the resultant vector with respect to the East direction:
$\tan \beta = \frac{v_{w} \sin(135^{\circ}) - v_{b} \sin(90^{\circ})}{v_{w} \cos(135^{\circ}) + v_{b} \cos(90^{\circ})}$ is not the standard approach. Let's use the vector components:
$\vec{v}_{w} = 72 \cos(45^{\circ}) \hat{i} + 72 \sin(45^{\circ}) \hat{j} = 50.91 \hat{i} + 50.91 \hat{j}$
$\vec{v}_{b} = 51 \hat{j}$
$\vec{v}_{wb} = \vec{v}_{w} - \vec{v}_{b} = 50.91 \hat{i} + (50.91 - 51) \hat{j} = 50.91 \hat{i} - 0.09 \hat{j}$
The angle $\theta$ with the East direction is $\tan \theta = \frac{|v_{wb, y}|}{|v_{wb, x}|} = \frac{0.09}{50.91} \approx 0.00177$.
$\theta \approx \tan^{-1}(0.00177) \approx 0.1^{\circ}$ South of East.
Thus,the flag flutters almost due East.