Concept Card: Relative Velocity in Two Dimensions (River-Boat & Rain-Man)
1. Fundamental Formulation of 2D Relative Velocity:
The velocity of object $A$ observed from the frame of reference of object $B$ is given by vector subtraction:
$\vec{v}_{AB} = \vec{v}_A – \vec{v}_B$
$\vec{v}_A = \vec{v}_{AB} + \vec{v}_B$
where $\vec{v}_A$ and $\vec{v}_B$ are measured relative to a common reference frame (the ground).
2. Rain-Man Problems:
- Let velocity of rain relative to ground be $\vec{v}_R = -v_R\hat{j}$ (falling vertically).
- Let velocity of man relative to ground be $\vec{v}_M = v_M\hat{i}$ (moving horizontally).
- Velocity of rain relative to the moving man ($\vec{v}_{RM}$):
$\vec{v}_{RM} = \vec{v}_R – \vec{v}_M = -v_M\hat{i} – v_R\hat{j}$ - Relative Speed of Rain:
$|\vec{v}_{RM}| = \sqrt{v_R^2 + v_M^2}$ - Umbrella Orientation:
To protect against rain, the umbrella must be oriented opposite to the relative velocity vector ($-\vec{v}_{RM} = v_M\hat{i} + v_R\hat{j}$):
$\tan\theta = \frac{|\text{Horizontal Component}|}{|\text{Vertical Component}|} = \frac{v_M}{v_R}$
where $\theta$ is the angle made by the umbrella with the vertical, tilted forward into the direction of motion. - Oblique / Wind-Blown Rain:
If rain falls with velocity $\vec{v}_R = -v_R\sin\alpha\hat{i} – v_R\cos\alpha\hat{j}$ and the man moves with $\vec{v}_M = v_M\hat{i}$:
$\vec{v}_{RM} = -(v_R\sin\alpha + v_M)\hat{i} – v_R\cos\alpha\hat{j}$
3. River-Boat (Swimmer-Stream) Problems:
Let river flow be along $+x$ with velocity $\vec{v}_R = v_r\hat{i}$.
Let river width be $d$ along $+y$.
Velocity of boat relative to water is $\vec{v}_{BR}$ (speed $v_b$).
Net velocity of boat relative to ground: $\vec{v}_B = \vec{v}_{BR} + \vec{v}_R$.
- Case A: Shortest Time of Crossing ($t_{\text{min}}$):
- To cross in minimum time, the cross-river component ($v_y$) must be maximized. The boat must head straight perpendicular to the flow ($\vec{v}_{BR} = v_b\hat{j}$).
- Minimum crossing time: $t_{\text{min}} = \frac{d}{v_b}$
- Downstream drift: $x_{\text{drift}} = v_r \cdot t_{\text{min}} = \left(\frac{v_r}{v_b}\right)d$
- Case B: Shortest Path / Zero Drift ($x_{\text{drift}} = 0$):
- Possible if and only if $v_b \ge v_r$ (boat speed $\ge$ river speed).
- The boat heads upstream at angle $\phi$ with the normal to the banks such that the horizontal component cancels river flow:
$v_b\sin\phi = v_r \implies \sin\phi = \frac{v_r}{v_b}$ - Time taken along shortest path:
$t = \frac{d}{v_b\cos\phi} = \frac{d}{\sqrt{v_b^2 – v_r^2}}$
- Case C: Minimum Drift when Swimmer is Slower than River ($v_b < v_r$):
- Zero drift is impossible. The minimum drift occurs when heading at angle $\phi$ with the normal where $\sin\phi = \frac{v_b}{v_r}$.
- Minimum drift: $x_{\text{min}} = d \sqrt{\frac{v_r^2}{v_b^2} – 1}$
4. Closest Approach / Collision in 2D:
In the reference frame of one object (say $B$ at the origin), object $A$ moves along the relative velocity line $\vec{v}_{AB}$. The minimum separation $d_{\text{min}}$ is the perpendicular dropped from $B$ onto the relative path line of $A$. Two objects collide if and only if the relative velocity points directly along their line of separation ($\vec{v}_{AB} \parallel -\vec{r}_{AB}$).
Solved Examples
Example 1 (Rain-Man Umbrella Dynamics):
Rain is falling vertically downwards with a speed of $30\text{ m/s}$. A cyclist rides horizontally East at $10\text{ m/s}$.
(a) At what angle with the vertical must the cyclist hold an umbrella to stay dry?
(b) If the cyclist speeds up to $30\text{ m/s}$, what should be the new umbrella angle?
(c) At what speed must the cyclist ride so that the raindrops appear to strike at $60^\circ$ to the vertical?
Solution:
(a) Rain velocity: $\vec{v}_R = -30\hat{j}\text{ m/s}$.
Cyclist velocity: $\vec{v}_C = +10\hat{i}\text{ m/s}$.
Relative velocity of rain: $\vec{v}_{RC} = \vec{v}_R – \vec{v}_C = -10\hat{i} – 30\hat{j}\text{ m/s}$.
$\tan\theta = \frac{v_C}{v_R} = \frac{10}{30} = \frac{1}{3} \implies \theta = \arctan(1/3) \approx 18.43^\circ$ with the vertical towards the East.
(b) At speed $v_C’ = 30\text{ m/s}$:
$\tan\theta’ = \frac{30}{30} = 1 \implies \theta’ = 45^\circ$ with the vertical towards the East.
(c) When angle with vertical is $60^\circ$:
$\tan(60^\circ) = \frac{v_C”}{v_R} \implies \sqrt{3} = \frac{v_C”}{30} \implies v_C” = 30\sqrt{3}\text{ m/s} \approx 51.96\text{ m/s}$.
Example 2 (River-Boat: Shortest Time vs. Zero Drift):
A river of width $d = 400\text{ m}$ flows due East at speed $v_r = 3.0\text{ m/s}$. A motorboat moves at speed $v_b = 5.0\text{ m/s}$ in still water.
(a) Find the heading direction and time taken to cross the river straight across to the opposite bank (zero drift).
(b) Find the shortest time required to cross the river and the resulting downstream drift.
Solution:
(a) For zero drift ($x_{\text{drift}} = 0$), the boat must steer upstream at angle $\phi$ with the normal (North):
$\sin\phi = \frac{v_r}{v_b} = \frac{3.0}{5.0} = 0.6 \implies \phi = 37^\circ$ West of North.
Effective cross-river velocity:
$v_y = \sqrt{v_b^2 – v_r^2} = \sqrt{5.0^2 – 3.0^2} = \sqrt{25 – 9} = \sqrt{16} = 4.0\text{ m/s}$.
Time taken along shortest path: $t = \frac{d}{v_y} = \frac{400\text{ m}}{4.0\text{ m/s}} = 100\text{ seconds}$.
(b) For shortest time, the boat must steer directly North ($\phi = 0^\circ$):
$t_{\text{min}} = \frac{d}{v_b} = \frac{400\text{ m}}{5.0\text{ m/s}} = 80\text{ seconds}$.
Downstream drift during this minimum time:
$x_{\text{drift}} = v_r \cdot t_{\text{min}} = 3.0\text{ m/s} \times 80\text{ s} = 240\text{ meters}$.
Example 3 (Minimum Drift when Swimmer is Slower than River):
A wide river flows at speed $v_r = 5.0\text{ m/s}$. A swimmer can swim at $v_s = 3.0\text{ m/s}$ in still water. The width of the river is $120\text{ m}$.
(a) Why is zero drift impossible?
(b) Find the steering angle with the river flow that produces the minimum possible downstream drift.
(c) Calculate the minimum drift.
Solution:
(a) Zero drift requires $\sin\phi = \frac{v_r}{v_s} = \frac{5}{3} > 1$, which has no real solution since $\sin\phi \le 1$. The river is faster than the swimmer, so the swimmer cannot prevent being swept downstream.
(b) Let the swimmer steer at angle $\theta$ with the river bank (upstream, so $\theta > 90^\circ$). Let $\phi = \theta – 90^\circ$ be the angle with the normal towards upstream.
$v_x = v_r – v_s\sin\phi = 5 – 3\sin\phi$.
$v_y = v_s\cos\phi = 3\cos\phi$.
Time of crossing: $t = \frac{d}{3\cos\phi}$.
Drift: $x = v_x t = \frac{d(5 – 3\sin\phi)}{3\cos\phi} = \frac{d}{3} [5\sec\phi – 3\tan\phi]$.
To minimize $x$, set $\frac{dx}{d\phi} = 0$:
$\frac{d}{d\phi}(5\sec\phi – 3\tan\phi) = 5\sec\phi\tan\phi – 3\sec^2\phi = 0 \implies 5\tan\phi = 3\sec\phi \implies 5\left(\frac{\sin\phi}{\cos\phi}\right) = \frac{3}{\cos\phi} \implies \sin\phi = \frac{3}{5} = \frac{v_s}{v_r}$.
Hence $\phi = \arcsin(3/5) = 37^\circ$ upstream from the normal, which corresponds to heading angle $\theta = 90^\circ + 37^\circ = 127^\circ$ with the downstream river flow.
(c) With $\sin\phi = \frac{3}{5}$, $\cos\phi = \frac{4}{5}$.
Minimum drift: $x_{\text{min}} = \frac{120}{3} \left[5\left(\frac{5}{4}\right) – 3\left(\frac{3}{4}\right)\right] = 40 \left[\frac{25 – 9}{4}\right] = 40 \left(\frac{16}{4}\right) = 40 \times 4 = 160\text{ meters}$.
Example 4 (Minimum Separation / Closest Approach Between Ships):
At $t = 0$, Ship $A$ is $10\text{ km}$ due North of Ship $B$. Ship $A$ steams due East at $20\text{ km/h}$, while Ship $B$ steams due North at $20\text{ km/h}$.
(a) Determine the relative velocity of Ship $A$ with respect to Ship $B$.
(b) Find the closest distance of approach between the two ships.
(c) Find the time after $t = 0$ when this closest approach occurs.
Solution:
(a) Velocities in Cartesian coordinates (East $= +\hat{i}$, North $= +\hat{j}$):
$\vec{v}_A = 20\hat{i}\text{ km/h}$
$\vec{v}_B = 20\hat{j}\text{ km/h}$
Relative velocity of $A$ with respect to $B$:
$\vec{v}_{AB} = \vec{v}_A – \vec{v}_B = 20\hat{i} – 20\hat{j}\text{ km/h}$.
Magnitude: $v_{\text{rel}} = \sqrt{20^2 + (-20)^2} = 20\sqrt{2}\text{ km/h}$.
Direction: $\theta = 45^\circ$ South of East.
(b) Position of $A$ relative to $B$ at $t = 0$: $\vec{r}_0 = 0\hat{i} + 10\hat{j}\text{ km}$.
Position vector at time $t$:
$\vec{r}(t) = \vec{r}_0 + \vec{v}_{AB} t = 10\hat{j} + (20\hat{i} – 20\hat{j})t = 20t\hat{i} + (10 – 20t)\hat{j}$.
Square of separation distance:
$D^2(t) = (20t)^2 + (10 – 20t)^2 = 400t^2 + 100 – 400t + 400t^2 = 800t^2 – 400t + 100$.
(c) Differentiating to find minimum separation:
$\frac{d(D^2)}{dt} = 1600t – 400 = 0 \implies t = \frac{400}{1600} = \frac{1}{4}\text{ hour} = 15\text{ minutes}$.
Closest distance of approach:
$D_{\text{min}}^2 = 800\left(\frac{1}{16}\right) – 400\left(\frac{1}{4}\right) + 100 = 50 – 100 + 100 = 50\text{ km}^2$.
$D_{\text{min}} = \sqrt{50} = 5\sqrt{2}\text{ km} \approx 7.07\text{ km}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
Rain is falling vertically downwards at $4\text{ m/s}$. A person walks horizontally with a speed of $3\text{ m/s}$. To keep from getting wet, the person must hold an umbrella at an angle $\theta$ with the vertical such that $\tan\theta$ is:
(A) $\frac{3}{4}$
(B) $\frac{4}{3}$
(C) $\frac{3}{5}$
(D) $\frac{4}{5}$
Problem 2 (JEE Main – Single Correct):
A river of width $1\text{ km}$ flows at $3\text{ km/h}$. A swimmer who can swim at $5\text{ km/h}$ in still water wants to cross directly opposite to the starting point. The time taken by the swimmer is:
(A) $12\text{ min}$
(B) $15\text{ min}$
(C) $20\text{ min}$
(D) $25\text{ min}$
Problem 3 (JEE Main – Single Correct):
A man running on a horizontal road at $v = 8\text{ km/h}$ finds the raindrops hitting him vertically. When he doubles his running speed to $16\text{ km/h}$, the raindrops hit him at an angle of $45^\circ$ with the vertical. The true speed of the raindrops is:
(A) $8\text{ km/h}$
(B) $8\sqrt{2}\text{ km/h}$
(C) $16\text{ km/h}$
(D) $12\text{ km/h}$
Problem 4 (JEE Main – Single Correct):
A boat crosses a river of width $d$ in the shortest possible time $t_1$. It crosses the same river along the shortest path in time $t_2$. If the boat’s speed in still water is $v_b$ and the river flow speed is $v_r$, the ratio $\frac{t_1}{t_2}$ is:
(A) $\sqrt{1 – \frac{v_r^2}{v_b^2}}$
(B) $\frac{1}{\sqrt{1 – v_r^2 / v_b^2}}$
(C) $1 – \frac{v_r}{v_b}$
(D) $\sqrt{1 + \frac{v_r^2}{v_b^2}}$
Problem 5 (JEE Main – Single Correct):
Car $A$ moves North at $30\text{ km/h}$ and Car $B$ moves East at $40\text{ km/h}$. The magnitude of the relative velocity of Car $A$ with respect to Car $B$ is:
(A) $10\text{ km/h}$
(B) $50\text{ km/h}$
(C) $70\text{ km/h}$
(D) $25\text{ km/h}$
Problem 6 (JEE Advanced – One or More Correct):
A boat can move with speed $v_b$ in still water, while the river flows at speed $v_r$ ($v_r > v_b$). Which of the following statements is/are correct?
(A) The boat can never reach a point directly opposite on the other bank.
(B) The minimum downstream drift occurs when the boat heads at an angle $\theta = \arcsin\left(\frac{v_b}{v_r}\right)$ with the normal to the river flow.
(C) The minimum time to cross the river is $\frac{d}{v_b}$.
(D) The trajectory of the boat in the river frame is a straight line.
Problem 7 (JEE Advanced – One or More Correct):
Rain is falling vertically at speed $v_R$. A man walks horizontally at speed $v_M$. Which of the following statements is/are TRUE?
(A) The relative speed of rain as perceived by the man is $\sqrt{v_R^2 + v_M^2}$.
(B) The faster the man moves horizontally, the greater the angle of his umbrella with the vertical.
(C) If the man accelerates forward, the relative velocity vector of rain continually tilts further from the vertical.
(D) If the man holds a vertical pipe while running, raindrops will enter and pass through the pipe without hitting the inner walls.
Problem 8 (JEE Advanced – One or More Correct):
Two bodies $A$ and $B$ are projected simultaneously into the air from different points on the ground with velocities $\vec{u}_A$ and $\vec{u}_B$. Which of the following statements is/are correct for their motion before either lands?
(A) The relative acceleration $\vec{a}_{AB}$ between them is identically zero.
(B) The relative trajectory of $A$ as seen by $B$ is a straight line.
(C) If their relative velocity vector is initially oriented along the line joining them, they will collide.
(D) The relative velocity vector rotates continuously due to gravity.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A swimmer wishes to cross a $500\text{ m}$ wide river flowing at $4.0\text{ m/s}$. The swimmer’s speed in still water is $5.0\text{ m/s}$. If the swimmer crosses along the shortest path (reaching directly opposite), the crossing time in seconds is $T$. Find $T$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A man standing on a road holds his umbrella at $30^\circ$ with the vertical to keep off the rain. He drops the umbrella and begins running at $10\text{ km/h}$. He finds that the raindrops are now hitting his head vertically. The true velocity of the rain relative to the ground in $\text{km/h}$ is $V_R$. Find the integer value of $V_R$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (B) | 4. (A) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C) | 8. (A, B, C) | 9. 167 (or 166.7) | 10. 20
Solution 1:
$\vec{v}_R = -4\hat{j}, \vec{v}_M = 3\hat{i} \implies \vec{v}_{RM} = -3\hat{i} – 4\hat{j}$.
$\tan\theta = \frac{|\text{Horizontal}|}{|\text{Vertical}|} = \frac{3}{4}$.
Correct Answer: (A)
Solution 2:
For zero drift, cross-river velocity $v_y = \sqrt{v_b^2 – v_r^2} = \sqrt{5^2 – 3^2} = 4\text{ km/h}$.
$t = \frac{d}{v_y} = \frac{1\text{ km}}{4\text{ km/h}} = 0.25\text{ h} = 15\text{ minutes}$.
Correct Answer: (B)
Solution 3:
Let true rain velocity be $\vec{v}_R = v_x\hat{i} – v_y\hat{j}$.
Case 1: When man runs at $8\hat{i}$, rain appears vertical $\implies (v_x – 8)\hat{i} – v_y\hat{j}$ is purely vertical $\implies v_x – 8 = 0 \implies v_x = 8\text{ km/h}$.
Case 2: When man runs at $16\hat{i}$, $\vec{v}_{RM} = (8 – 16)\hat{i} – v_y\hat{j} = -8\hat{i} – v_y\hat{j}$.
Angle is $45^\circ \implies \tan(45^\circ) = \frac{8}{v_y} = 1 \implies v_y = 8\text{ km/h}$.
True speed of rain: $v_R = \sqrt{v_x^2 + v_y^2} = \sqrt{8^2 + 8^2} = 8\sqrt{2}\text{ km/h}$.
Correct Answer: (B)
Solution 4:
$t_1 = \frac{d}{v_b}$.
$t_2 = \frac{d}{\sqrt{v_b^2 – v_r^2}} = \frac{d}{v_b\sqrt{1 – v_r^2 / v_b^2}}$.
Ratio $\frac{t_1}{t_2} = \frac{d/v_b}{\frac{d}{v_b\sqrt{1 – v_r^2 / v_b^2}}} = \sqrt{1 – \frac{v_r^2}{v_b^2}}$.
Correct Answer: (A)
Solution 5:
$\vec{v}_A = 30\hat{j}, \vec{v}_B = 40\hat{i}$.
$\vec{v}_{AB} = \vec{v}_A – \vec{v}_B = -40\hat{i} + 30\hat{j}$.
$|\vec{v}_{AB}| = \sqrt{(-40)^2 + (30)^2} = \sqrt{1600 + 900} = 50\text{ km/h}$.
Correct Answer: (B)
Solution 6:
All four statements (A, B, C, D) are classic established results for river crossing when swimmer velocity is less than river flow speed.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: $|\vec{v}_{RM}| = \sqrt{v_R^2 + v_M^2}$.
– (B) True: $\tan\theta = \frac{v_M}{v_R}$ increases with $v_M$.
– (C) True: If $v_M$ increases, $\theta$ increases.
– (D) False: To pass through without hitting walls, the pipe must be tilted along the relative velocity $\vec{v}_{RM}$.
Correct Answer: (A, B, C)
Solution 8:
– (A) True: $\vec{a}_{AB} = (-g\hat{j}) – (-g\hat{j}) = 0$.
– (B) True: Since relative acceleration is zero, relative trajectory is a straight line.
– (C) True: Condition of direct collision under zero relative acceleration.
– (D) False: Relative velocity $\vec{v}_{AB} = \vec{u}_A – \vec{u}_B$ is completely independent of time.
Correct Answer: (A, B, C)
Solution 9:
Effective cross-river speed: $v_y = \sqrt{5.0^2 – 4.0^2} = \sqrt{25 – 16} = 3.0\text{ m/s}$.
Time taken: $T = \frac{d}{v_y} = \frac{500\text{ m}}{3.0\text{ m/s}} = \frac{500}{3} \approx 166.7\text{ s}$ (or $167\text{ s}$).
Correct Answer: 167 (or 166.7)
Solution 10:
When standing, the umbrella is at $30^\circ$ to vertical $\implies$ rain falls at $30^\circ$ to vertical:
$\vec{v}_R = -v_R\sin(30^\circ)\hat{i} – v_R\cos(30^\circ)\hat{j}$.
When running at $v_M = 10\hat{i}$, rain hits vertically $\implies$ horizontal relative velocity is zero:
$v_R\sin(30^\circ) = v_M = 10\text{ km/h} \implies v_R\left(\frac{1}{2}\right) = 10 \implies v_R = 20\text{ km/h}$.
Correct Answer: 20