Relative Velocity: River Crossing, Wind and Rain Problems

Motion Depends on Who Is Watching

Sitting in a bus moving at 60 km/h, the passenger beside you seems perfectly still, yet to someone standing at the roadside, both of you are racing past. A car overtaking you at 70 km/h seems to creep forward slowly. Rain falling straight down seems to slant towards you when you run. All these are examples of relative velocity: the velocity of one object as seen by an observer who may also be moving. This tutorial covers relative velocity in one and two dimensions, the classic river-crossing and aircraft-in-wind problems, and rain and umbrellas, with worked examples of the kind set in WAEC, JAMB and PHY 101. It builds on vectors and scalars.

The Key Idea

The velocity of object A relative to object B is the velocity A appears to have to an observer moving with B:

v(A relative to B) = v(A) − v(B)

Here v(A) and v(B) are both measured relative to the ground. Because velocities are vectors, this is vector subtraction, which must take direction into account. A useful check is that v(B relative to A) = −v(A relative to B): it has the same size but the opposite direction.

Relative Velocity in One Dimension

Along a straight line, choose one direction as positive.

Same direction: subtract the speeds. If car A moves at 80 km/h and car B at 60 km/h in the same direction, A moves away from B at 80 − 60 = 20 km/h.
Opposite directions: add the speeds. If they move towards each other, A approaches B at 80 + 60 = 140 km/h.

Worked example 1: overtaking

A car 4 m long, travelling at 25 m/s, overtakes a lorry 16 m long travelling at 20 m/s in the same direction. How long does the overtaking take, from when the car's front reaches the lorry's back to when the car's back passes the lorry's front?

The relative velocity of the car is 25 − 20 = 5 m/s. The car must gain 4 + 16 = 20 m on the lorry, so the time is 20 ÷ 5 = 4 s.

Worked example 2: trains passing

Two trains, each 150 m long, travel towards each other on parallel tracks at 20 m/s and 30 m/s. How long do they take to pass each other completely?

The relative speed is 20 + 30 = 50 m/s, and the total distance is 150 + 150 = 300 m, so the time is 300 ÷ 50 = 6 s.

Relative Velocity in Two Dimensions

When the velocities are not along the same line, use vector addition and subtraction. For perpendicular velocities, use Pythagoras' theorem for the size and trigonometry (usually tan) for the direction.

Worked example 3: two cars at right angles

Car A travels east at 30 m/s and car B travels north at 20 m/s. Find the velocity of B relative to A.

v(B relative to A) = v(B) − v(A) = (20 m/s north) + (30 m/s west), because subtracting an eastward velocity is the same as adding a westward one.

Size = √(20² + 30²) = √1300 ≈ 36.1 m/s.

Direction: tan θ = 30 ÷ 20 = 1.5, so θ ≈ 56.3° west of north, a bearing of about 304°. To the driver of car A, car B seems to be moving north-west, even though it is actually travelling north.

River-Crossing Problems

A boat moves through the water, and the water moves relative to the bank. The boat's velocity relative to the bank is the vector sum:

v(boat relative to bank) = v(boat relative to water) + v(water relative to bank)

Worked example 4: heading straight across

A boat that can move at 4 m/s in still water heads straight across a river 100 m wide. The current flows at 3 m/s. Find the time to cross, the resultant velocity and how far downstream the boat lands.

Time: only the 4 m/s across the river carries the boat across, so t = 100 ÷ 4 = 25 s. The current does not change the crossing time.
Resultant speed: √(4² + 3²) = 5 m/s, at an angle of tan⁻¹(3/4) ≈ 36.9° downstream from the straight-across direction.
Drift downstream: 3 m/s × 25 s = 75 m.

Worked example 5: landing directly opposite

In which direction must the same boat head to land directly opposite its starting point, and how long does it take?

The boat must point partly upstream so that the upstream part of its velocity cancels the current exactly: 4 sin θ = 3, so sin θ = 0.75 and θ ≈ 48.6° upstream from the straight-across direction.

Its speed straight across is then √(4² − 3²) = √7 ≈ 2.65 m/s, so the time is 100 ÷ 2.65 ≈ 37.8 s, longer than before.

This is only possible if the boat's speed in still water is greater than the current's speed. Otherwise, it will always be carried downstream.

The shortest time versus the shortest path

For the shortest time, head straight across.
For the shortest path (straight across), head upstream at the angle that cancels the current.

Aircraft and Wind

Aircraft problems work the same way: the plane's airspeed is its velocity relative to the air, and the wind moves the air relative to the ground. The plane's ground velocity is their vector sum.

Worked example 6

A plane flies with an airspeed of 200 km/h heading due north, while a wind blows from west to east at 50 km/h. Find its velocity relative to the ground.

Ground speed = √(200² + 50²) = √42,500 ≈ 206.2 km/h.

Direction: tan θ = 50 ÷ 200 = 0.25, so θ ≈ 14.0° east of north, a bearing of 014°.

To fly due north over the ground, the pilot would head slightly west of north, into the wind, just as the boat heads upstream.

Rain and Umbrellas

Rain falls vertically at 8 m/s on a windless day, and a person walks at 6 m/s. Relative to the walker, the rain has a horizontal velocity of 6 m/s towards them, as well as its 8 m/s downwards.

Relative speed of the rain = √(8² + 6²) = 10 m/s, at an angle tan⁻¹(6/8) ≈ 36.9° to the vertical, coming from in front.

So the walker should tilt the umbrella forward by about 37°. This is why rain seems to slant towards you when you run, and why the front of a moving car's windscreen gets wetter than the back window.

A Step-by-Step Method

Write down each velocity with its direction, relative to the ground.
Decide what is wanted: "A relative to B" means v(A) − v(B).
Draw a vector diagram. For subtraction, reverse the direction of the vector being subtracted and add it.
Use Pythagoras for perpendicular vectors, or resolve into components (or use the cosine rule) for other angles.
Give both the size and the direction (an angle or a bearing).

Common Mistakes with Relative Velocity

Adding speeds when objects move in the same direction. Subtract them.
Forgetting direction in two-dimensional problems, and giving only a speed.
Thinking the current changes the crossing time when the boat heads straight across. It only changes where the boat lands.
Mixing up the heading and the actual path. The boat points one way but moves another.
Getting "A relative to B" backwards. It is v(A) − v(B), which is how A looks to an observer moving with B.

Relative Velocity: Practice Questions

A motorcycle travelling at 90 km/h overtakes a bus travelling at 70 km/h on a straight road. What is the velocity of the motorcycle relative to the bus?
Two cars approach each other at 15 m/s and 25 m/s. What is their relative speed?
A swimmer who can swim at 1.5 m/s in still water heads straight across a 60 m wide river flowing at 2 m/s. How long does the crossing take, and how far downstream does the swimmer land?
Can the swimmer in question 3 reach the point directly opposite? Explain.
A plane with an airspeed of 300 km/h heads due east in a wind of 40 km/h blowing due south. Find its ground speed.
Rain falls vertically at 12 m/s. A cyclist rides at 5 m/s. Find the speed of the rain relative to the cyclist.

Answers: 1. 20 km/h in the direction of motion. 2. 40 m/s. 3. t = 60 ÷ 1.5 = 40 s, and the drift is 2 × 40 = 80 m. 4. No. The current (2 m/s) is faster than the swimmer (1.5 m/s), so the swimmer can never cancel it. 5. √(300² + 40²) ≈ 302.7 km/h. 6. √(12² + 5²) = 13 m/s.

Conclusion

Relative velocity is the velocity of one object as seen from another: v(A relative to B) = v(A) − v(B). In one dimension, subtract speeds for objects moving the same way and add them for objects moving towards each other. In two dimensions, combine vectors using Pythagoras and trigonometry. The same idea solves river crossings (head straight across for the shortest time, upstream for the shortest path), aircraft flying in wind, and the slant of falling rain.