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Relative velocity in one dimension
Relative velocity compares two objects moving along the same line:
\[v_{AB} = v_A - v_B\]
This reads as “the velocity of A relative to B”.
\(v_{AB}\) means “velocity of A as observed from B” or “velocity of A relative to B”
Interactive: One-Dimensional Relative Motion
Consider two cars on a highway. Their relative velocity determines how quickly the gap between them changes.
Example: Car A at 25 m/s, Car B at 18 m/s (both east)
- Relative velocity: \(v_{AB} = 25 - 18 = 7\) m/s
- Car A approaches Car B at 7 m/s
Relative velocity in two dimensions
Vectors are subtracted component-wise. The relative velocity points from the observer to the object being described.
\[\vec{v}_{AB} = \vec{v}_A - \vec{v}_B\]
Interactive: Boat and River Current
A classic relative motion problem: a boat crossing a river with current.
Interpreting the diagram:
- Blue vector: boat’s velocity relative to water (heading north)
- Green vector: water’s velocity relative to ground (current flowing east)
- Red dashed vector: boat’s velocity relative to ground (resultant)
Reference frames
A statement about motion is incomplete without a frame. The same motion can be described differently in different frames without contradiction.
All velocity measurements require specifying:
- What is moving
- Relative to what it’s being measured
Common Relative Motion Scenarios
| Overtaking cars |
Ground |
Slower car |
Difference in speeds |
| Boat in current |
Water |
Ground |
Vector sum |
| Rain on cyclist |
Ground |
Cyclist |
Vector difference |
| Aircraft in wind |
Air |
Ground |
Vector sum |
Worked Examples
Example 1: Overtaking cars
Car A travels at 25 m/s east. Car B travels at 18 m/s east. Find A relative to B.
Solution:
- \(v_{AB} = v_A - v_B = 25 - 18 = 7\) m/s
- The positive result indicates A moves east relative to B
- A closes the gap at 7 m/s
Example 2: Boat and current
A boat heads due north at 4.0 m/s relative to the water. The current is 1.5 m/s east. Find velocity relative to the ground.
Solution:
- Components: \(v_x = 1.5\) m/s, \(v_y = 4.0\) m/s
- Speed: \(\sqrt{1.5^2 + 4.0^2} = \sqrt{18.25} = 4.3\) m/s
- Direction: \(\tan^{-1}(1.5/4.0) = 20°\) east of north
Example 3: Rain and a moving cyclist
Rain falls vertically at 6.0 m/s. A cyclist rides east at 5.0 m/s. Find the rain velocity relative to the cyclist.
Solution:
The rain’s velocity relative to the cyclist is found by vector subtraction:
\[\vec{v}_{rain/cyclist} = \vec{v}_{rain} - \vec{v}_{cyclist}\]
- Rain (relative to ground): \(v_x = 0\), \(v_y = -6.0\) m/s (downward)
- Cyclist (relative to ground): \(v_x = 5.0\) m/s, \(v_y = 0\)
- Rain relative to cyclist: \(v_x = 0 - 5.0 = -5.0\) m/s, \(v_y = -6.0\) m/s
Result: - Speed: \(\sqrt{5.0^2 + 6.0^2} = 7.8\) m/s - Direction: \(\tan^{-1}(5.0/6.0) = 40°\) from vertical, toward the cyclist (west of vertical)
The rain appears to come from ahead! This is why cyclists lean forward in rain.
Example 4: Head-on collision approach
Two trains approach each other. Train A travels at 30 m/s east, Train B at 25 m/s west.
Solution:
Taking east as positive: - \(v_A = +30\) m/s - \(v_B = -25\) m/s
Relative velocity: - \(v_{AB} = v_A - v_B = 30 - (-25) = 55\) m/s
The trains approach each other at the sum of their speeds.
Interactive: Aircraft Navigation
An aircraft must aim off-course to compensate for wind:
Navigation problem: To fly due north (90°), the pilot must aim slightly west to compensate for the eastward wind.