Motion in a Straight Line · Lesson 8 of 10
Relative Motion and Meeting Problems
“Describe motion from different observers and solve overtaking, approaching and relative impact-speed problems.”
• Derive relative position and relative velocity in one dimension. • Use signed velocities in a common reference frame. • Calculate meeting and overtaking times. • Distinguish speed relative to a moving source, the ground and a target.
Motion depends on the observer
A person seated in a train has zero velocity relative to the train but shares the train’s velocity relative to the ground. A nearby train may appear to slide backward even while both trains move forward relative to the ground. To prevent ambiguity, label each velocity with the observer it is measured relative to. In ordinary nonrelativistic motion, velocities in one direction combine through signed addition and subtraction.
Imagine shifting the coordinate origin to observer A at every instant. The coordinate assigned to B is its position minus A’s position. Differentiating this difference gives the relative velocity. This derivation is more reliable than memorising “subtract for same direction, add for opposite direction”, because signs automatically account for the directions.
Same-direction and opposite-direction motion
Problem
Vehicle A moves east at 20 m/s; B moves east at 15 m/s. Find each velocity relative to the other.
- 1.Take east positive. Relative velocity of B with respect to A is 15 − 20 = −5 m/s.
- 2.A sees B move west at 5 m/s even though B travels east relative to the road.
- 3.Conversely, v_A/B = +5 m/s. Equal ground velocities would give zero relative velocity and constant separation.
Problem
A travels east at 20 m/s and B travels west at 15 m/s. Initially B is 350 m east of A. When do they meet?
- 1.Choose x_A(0) = 0 and x_B(0) = 350 m. Then v_A = +20 m/s and v_B = −15 m/s.
- 2.The signed separation is x_B − x_A = 350 + (−15 − 20)t = 350 − 35t.
- 3.At meeting, separation is zero, giving t = 10 s. The closing speed is 35 m/s, the magnitude of the relative velocity.
Deriving the meeting condition
If the initial separation is positive and relative velocity is also positive, B is pulling farther ahead, so no future meeting occurs under those unchanged velocities. A negative algebraic meeting time describes a meeting in the backward extension of the motion. If relative velocity is zero, separation stays fixed: distinct starting positions never meet, while identical starting positions stay together in this model.
Problem
A cyclist at 6 m/s starts 100 m behind a walker travelling in the same direction at 1 m/s. Find catch time and place.
- 1.Put the cyclist at x = 0 and the walker at x = 100 m at t = 0.
- 2.Write x_c = 6t and x_w = 100 + t. Setting them equal gives 5t = 100, so t = 20 s.
- 3.The meeting position is x = 120 m. The cyclist covers 120 m; the walker covers 20 m. The 100 m gap is closed at 5 m/s.
Changing reference frame in a projectile-speed question
A launch speed relative to a moving vehicle is not automatically the launch speed relative to the road. First transform it into the ground frame using signed addition. Then subtract the target’s ground velocity to get the projectile’s velocity relative to the target. The relative speed is the relevant kinematic speed at impact; actual damage also depends on mass, materials and collision details.
Problem
A van moves at 30 km/h and launches a projectile forward at 150 m/s relative to the van. A target ahead moves in the same direction at 192 km/h. Find projectile speed relative to the target, neglecting changes during flight.
- 1.Choose forward positive. Van ground velocity = 30/3.6 = 8.333… m/s; target ground velocity = 192/3.6 = 53.333… m/s.
- 2.Projectile ground velocity = 150 + 8.333… = 158.333… m/s.
- 3.Projectile velocity relative to target = 158.333… − 53.333… = +105 m/s. The relative speed is 105 m/s.
- 4.Subtracting the target speed directly from 150 m/s would mix two reference frames and give the wrong answer.
When acceleration matters
Relative motion is not restricted to constant velocities. If A and B each have constant acceleration, subtract their position equations: relative position = initial separation + initial relative velocity × t + ½(a_B − a_A)t². Equal accelerations cancel, so relative velocity remains constant even if both ground velocities change. If accelerations differ, solve the resulting quadratic and check which root represents the first relevant encounter. The relative definitions remain valid, but a constant-speed gap formula no longer suffices.
Quiz
A train moves east at 12 m/s. A passenger walks west at 2 m/s relative to the train. What is the passenger’s ground velocity?
Key Takeaways
• State the observer for every velocity. • Relative velocity is object velocity minus observer velocity in one common frame. • Meeting means equal positions at the same time. • Use signed velocities so approaching and overtaking cases follow the same equation.