Subtract the Observer Velocity, Then Solve in the Observer Frame
Relative-motion questions look different (rain, boats, trains, escalators, ships) but all use one move: the velocity of A as seen by B is v_A - v_B. Subtract the observer's velocity first, then solve in the observer's frame.
Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.
Rain, running men and tilted tubes
The rain velocity seen by a man is v_rain - v_man. If rain falls with velocity (2, -4) m/s and a man must see it hit at 5 m/s, then (2 - vx)² + 16 = 25, so vx = -1 or vx = 5: either -i or 5i works. For a boy running at v while rain falls vertically at u, the rain relative to him slants backward with horizontal component v and vertical component u, so he must tilt a tube forward by tan⁻¹(v/u) from the vertical for the drops to enter without touching its sides. A stone released from a train window has the train's horizontal velocity at the moment of release, so the ground observer sees it follow a parabola and the train's passenger sees it fall almost vertically.
Boats, trains and escalators
Upstream the effective speed is (v_boat - v_river) and downstream it is (v_boat + v_river). A boat rowing at 4 km/h in a 2 km/h river goes 2 km upstream in 1 h and returns in 20 min, so the round trip takes 80 min. For two trains of length 100 m, overtaking uses the difference of speeds and crossing uses the sum, with the distance equal to the total length 200 m: 200/20 = 10 m/s and 200/10 = 20 m/s, so the speeds are 15 m/s and 5 m/s.
Overtaking and bus-service problems
To find when a runner reaches a moving target, write both position equations and equate them. For a man running at 4 m/s who is 6 m behind the door of a bus that starts from rest with acceleration 1.2 m/s², 4t = 6 + 0.6 t², which gives t = 2.3 s (the earlier root) and a second root of about 4.4 s that does not matter. For a cyclist at 20 km/h who meets buses every 18 min in the direction of motion and every 6 min the other way, the bus spacing d satisfies d = (v - 20)(18/60) = (v + 20)(6/60), so v = 40 km/h, d = 6 km and the period is d / v = 9 min.
Closest approach and collision
Work in the frame of one object, so the other has a relative position r and a constant relative velocity v_rel.
A ship moving east at 10 km/h with a second ship 100 km north moving south at 10 km/h gives a relative velocity of magnitude 10√2 at 45 degrees, and the closest approach is after 5 h. Two particles collide only if the relative position vector points along the relative velocity: (r1 - r2)/|r1 - r2| = (v2 - v1)/|v2 - v1|.
OlympIQ exam lens
- ✓Velocity of A relative to B = v_A - v_B, as a vector subtraction
- ✓Rain tilt: tan of the angle from the vertical = (observer speed) / (rain speed)
- ✓Upstream speed = v_boat - v_river; downstream = v_boat + v_river; overtaking uses a difference, crossing uses a sum
- ✓Moving walkways and escalators: rates add, 1/t = 1/t1 + 1/t2
- ✓Closest approach is at t = -(r . v_rel)/|v_rel|²; collision needs r along -v_rel
Do not add or subtract speeds without checking directions. Relative velocity is a vector difference, so objects moving in the same direction use a difference of speeds and objects moving oppositely use a sum, and perpendicular motions need Pythagoras.
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