Relative Velocity – Boats & Trains Problems

1. Introduction

In physics, velocity is not absolute—it is always measured relative to a frame of reference. This is the core idea behind relative velocity.

For example, if you are sitting inside a moving train and toss a ball straight up, the ball seems to move vertically up and down to you. But for someone standing on the ground, the ball follows a parabolic path because it also has horizontal velocity equal to that of the train.

This difference in perception is exactly what relative velocity is about.

Relative velocity is extremely useful in solving problems involving boats and streams (rivers), trains, cars, and even airplanes flying against or with wind currents.

In this post, we’ll deeply study:

Definition and formula of relative velocity
Graphical/vector approach
Boats and streams problems
Trains and cars problems
Many solved examples
Real-life applications

2. Relative Velocity – Concept & Formula

Definition

The velocity of an object A relative to another object B is the velocity that object A appears to have when observed from B’s frame of reference.

Formula

V⃗A/B=V⃗A−V⃗B\vec{V}_{A/B} = \vec{V}_A – \vec{V}_BVA/B=VA−VB

Where:

V⃗A/B\vec{V}_{A/B}VA/B = velocity of A relative to B
V⃗A\vec{V}_AVA = velocity of A relative to ground
V⃗B\vec{V}_BVB = velocity of B relative to ground

👉 This is a simple vector subtraction.

Example 1: Two Cars

Car A moves at 60 km/h east.
Car B moves at 40 km/h east.

Velocity of A relative to B: VA/B=60−40=20 km/h (east)V_{A/B} = 60 – 40 = 20 \, \text{km/h (east)}VA/B=60−40=20km/h (east)

👉 For a passenger in car B, car A seems to move 20 km/h east.

Example 2: Cars in Opposite Direction

Car A moves at 60 km/h east.
Car B moves at 40 km/h west.

VA/B=60−(−40)=100 km/hV_{A/B} = 60 – (-40) = 100 \, \text{km/h}VA/B=60−(−40)=100km/h

👉 For a passenger in car B, car A appears to move at 100 km/h.

3. Relative Velocity in Two Dimensions

When velocities are not along the same line (like boats crossing rivers), we use vector resolution.

Steps:

Draw velocity vectors.
Use vector subtraction: V⃗A/B=V⃗A−V⃗B\vec{V}_{A/B} = \vec{V}_A – \vec{V}_BVA/B=VA−VB.
Resolve into components (x and y).
Use Pythagoras and trigonometry to find magnitude and direction.

4. Boats and Streams Problems

Concept

A boat has a velocity relative to still water = VbV_bVb.
The river has velocity relative to ground = VrV_rVr.

The boat’s resultant velocity relative to ground is: V⃗bg=V⃗b+V⃗r\vec{V}_{bg} = \vec{V}_b + \vec{V}_rVbg=Vb+Vr

Depending on the situation:

If the boat goes downstream (with current): V=Vb+VrV = V_b + V_rV=Vb+Vr
If the boat goes upstream (against current): V=Vb−VrV = V_b – V_rV=Vb−Vr
If the boat wants to cross river perpendicular to current, it must adjust its direction.

Case 1: Boat Going Downstream

Speed of boat relative to ground: Vdown=Vb+VrV_{down} = V_b + V_rVdown=Vb+Vr

Time to cover distance ddd: t=dVb+Vrt = \frac{d}{V_b + V_r}t=Vb+Vrd

Case 2: Boat Going Upstream

Speed of boat relative to ground: Vup=Vb−VrV_{up} = V_b – V_rVup=Vb−Vr

Time to cover distance ddd: t=dVb−Vrt = \frac{d}{V_b – V_r}t=Vb−Vrd

👉 Note: Boat can move upstream only if Vb>VrV_b > V_rVb>Vr.

Case 3: Boat Crossing the River Perpendicular

If the boat is aimed perpendicular to the riverbank, the current will push it sideways.

Effective velocity across river = VbV_bVb.
Drift along river = Vr×tV_r \times tVr×t.
Time to cross = widthVb\frac{width}{V_b}Vbwidth.

Case 4: Boat Reaching Exactly Opposite Point

To avoid drifting, the boat must be pointed at an angle upstream.

Condition: Vbsin⁡θ=VrV_b \sin \theta = V_rVbsinθ=Vr

Where θ\thetaθ is the angle boat must make with perpendicular.

Time to cross = t=widthVbcos⁡θt = \frac{width}{V_b \cos \theta}t=Vbcosθwidth

Solved Example (Boat)

A river is 200 m wide, current speed = 3 m/s, boat speed in still water = 5 m/s.

If boat heads perpendicular to river:

Time to cross: 200/5=40 s200 / 5 = 40 \, s200/5=40s.
Drift: 3×40=120 m3 \times 40 = 120 \, m3×40=120m.

If boat wants to go straight across (no drift):

sin⁡θ=35=0.6⇒θ=37∘\sin \theta = \frac{3}{5} = 0.6 \quad \Rightarrow \theta = 37^\circsinθ=53=0.6⇒θ=37∘

Effective velocity across = Vbcos⁡θ=5×0.8=4 m/sV_b \cos \theta = 5 \times 0.8 = 4 \, m/sVbcosθ=5×0.8=4m/s.
Time = 200/4=50 s200/4 = 50 \, s200/4=50s.

5. Trains Problems

Relative velocity is very important in train problems.

Case 1: Trains Moving in Same Direction

Relative speed = difference of speeds.

Example:

Train A = 80 km/h
Train B = 60 km/h
Relative speed = 20 km/h20 \, km/h20km/h.

If Train A is 200 m long, time to cross Train B = t=20020×10003600=36 st = \frac{200}{20 \times \frac{1000}{3600}} = 36 \, st=20×36001000200=36s

Case 2: Trains Moving in Opposite Directions

Relative speed = sum of speeds.

Example:

Train A = 90 km/h
Train B = 60 km/h
Relative speed = 150 km/h=41.67 m/s150 \, km/h = 41.67 \, m/s150km/h=41.67m/s.

If length of each train = 150 m, total distance to cover = 300 m.

Time = 300/41.67≈7.2 s300 / 41.67 \approx 7.2 \, s300/41.67≈7.2s.

Case 3: Train Passing a Platform

Relative speed = speed of train.

If train length = 200 m, platform length = 300 m, train speed = 72 km/h = 20 m/s.

Time = Distance / Speed = (200+300)/20 = 25 s.

Case 4: Train Passing a Man

Relative speed = speed of train – speed of man (if both moving same direction).

Example:
Train = 72 km/h = 20 m/s
Man walking = 2 m/s in same direction
Relative speed = 18 m/s

If train length = 180 m, time to cross man = 180/18 = 10 s.

6. Important Tips for Boats & Trains Problems

Always convert km/h to m/s when required:

1 km/h=518 m/s1 \, km/h = \frac{5}{18} \, m/s1km/h=185m/s

Carefully check direction: same or opposite.
For boats, always compare VbV_bVb and VrV_rVr.
If solving for minimum time, set boat direction straight across.
If solving for shortest distance, set boat direction at an angle to cancel drift.

7. Common Mistakes Students Make

Forgetting to take relative velocity (subtracting instead of adding or vice versa).
Using wrong conversion of units.
Mixing up “minimum time” vs. “shortest path” conditions in river problems.
Forgetting to add train lengths when calculating crossing time.

8. Real-Life Applications

Boats crossing rivers.
Airplanes flying with/against wind.
Trains and cars overtaking each other.
Cyclists running in opposite directions in races.
Astronauts docking spacecraft in orbit (a real-life high-tech application of relative velocity).

9. Practice Problems

A boat can row at 6 km/h in still water. If the stream flows at 2 km/h, find the time taken to row 8 km downstream and return.
Two trains of lengths 120 m and 100 m move in opposite directions at 54 km/h and 36 km/h. Find the time they take to cross each other.
A 150 m long train passes a man walking at 5 km/h in the same direction in 15 seconds. Find the speed of the train.
A river is 300 m wide, current = 2 m/s, boat speed = 4 m/s. Find the angle at which boat must be rowed to reach directly opposite.