Relative Position, Velocity, and Acceleration
For particles A and B in a nonrotating frame, r_AB = r_A - r_B, v_AB = v_A - v_B, and a_AB = a_A - a_B. Closest approach occurs when relative position is perpendicular to relative velocity.
Why this shows up in the exam
Predicting ship separation · Finding collision conditions · Converting motion between translating frames
Learn the idea
Motion seen from B is found by subtracting B's motion from A's motion. Two observers can assign different velocities to the same object. Subtracting the observer's velocity removes the motion carried by that frame and leaves the closing or separating motion.
🧠 Memory hook: Relative means subtract the observer.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- v_AB = v_A - v_B — velocity of A relative to B
- r_AB(t) = r_AB(0) + v_AB t — relative position for constant relative velocity
- t_min = -(r₀ dot v_rel)/|v_rel|² — time of closest approach when it is nonnegative
How to approach it
- 1Name the observed object first and observer second
- 2Subtract their vectors in that order
- 3Use relative position to test meeting or minimum distance
Common slip-ups that cost marks
- •Reversing A relative to B
- •Adding velocities without a sign diagram
- •Using closest-approach time when relative acceleration is nonzero
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?
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