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Physics · Ch 2 — Kinematics

Relative Velocity in One and Two Dimensional Motion

2.10.2

Relative Velocity in One and Two Dimensional Motion

When two objects A and B move with their own velocities, what an observer riding along with B actually perceives is not A's velocity by itself, but the relative velocity of A with respect to B — the rate at which A's position changes as seen from B. It is defined as the vector difference of the two velocities:

v⃗AB=v⃗A−v⃗Band symmetricallyv⃗BA=v⃗B−v⃗A=−v⃗AB\vec v_{AB} = \vec v_A - \vec v_B \qquad\text{and symmetrically}\qquad \vec v_{BA} = \vec v_B - \vec v_A = -\vec v_{AB}

Case 1 — same direction. If A and B move along parallel straight tracks in the same direction with speeds vAv_A and vBv_B (measured with respect to the ground), then vAB=vA−vBv_{AB} = v_A - v_B and vBA=vB−vAv_{BA} = v_B - v_A: the magnitude of the relative velocity of one with respect to the other equals the difference of their speeds. (E.g. car A at 35 km h−1^{-1} east and car B at 40 km h−1^{-1} east: to a passenger in A, car B appears to creep ahead eastward at 5 km h−1^{-1}; to a passenger in B, car A appears to fall behind, i.e. drift westward, at 5 km h−1^{-1}.)

Case 2 — opposite directions. If A and B move along the same track but in opposite directions, then their relative velocity is vAB=vA−(−vB)=vA+vBv_{AB}=v_A-(-v_B)=v_A+v_B: the magnitude of the relative velocity equals the sum of their speeds.

Case 3 — general angle θ\theta between v⃗A\vec v_A and v⃗B\vec v_B. Using the law-of-cosines result from vector subtraction (§2.3.4),

vAB=vA2+vB2−2vAvBcos⁡θ,tan⁡β=vBsin⁡θvA−vBcos⁡θv_{AB} = \sqrt{v_A^2+v_B^2-2v_Av_B\cos\theta}, \qquad \tan\beta = \frac{v_B\sin\theta}{v_A-v_B\cos\theta}

(with β\beta the angle between v⃗AB\vec v_{AB} and v⃗B\vec v_B). This reduces to Case 1 when θ=0°\theta=0° (vAB=vA−vBv_{AB}=v_A-v_B, along v⃗A\vec v_A; and correspondingly vBA=vB+vAv_{BA}=v_B+v_A along v⃗B\vec v_B) and to Case 2 when θ=180°\theta=180° (vAB=vA+vBv_{AB}=v_A+v_B, along v⃗A\vec v_A; vBA=vB−vAv_{BA}=v_B-v_A along v⃗B\vec v_B). For two bodies moving at right angles (θ=90°\theta=90°), vAB=vA2+vB2v_{AB}=\sqrt{v_A^2+v_B^2}. …

Figure 2.36Angle of the umbrella with respect to falling rain

What this figure shows. A person walking horizontally with velocity v⃗M\vec v_M while rain falls vertically with velocity v⃗R\vec v_R; the vector construction v⃗RM=v⃗R−v⃗M\vec v_{RM} = \vec v_R - \vec v_M (obtained by adding v⃗R\vec v_R and −v⃗M-\vec v_M) gives the rain's velocity as seen by the walker, tilted at angle θ\theta from the vertical — the angle at which the umbrella must be tilted …