← Calculus methods

Catch-up and relative-motion problems

A kinematics scenario one level beyond a single moving object: two objects, two separate equations of motion, and a question about when (or whether) they meet.

The method for a two-object catch-up problem

  • 1. Write a SEPARATE displacement equation for each object, using the SAME variable for time and a SHARED reference point (e.g. the dock, or the starting line).
  • 2. If one object has a head start, build that directly into its displacement equation as a starting offset.
  • 3. The two objects meet when their displacements are EQUAL — set the two expressions equal to each other and solve for t.
  • 4. Substitute that t back into either displacement equation to find WHERE they meet, not just when.
Worked example A boat catching up to a cruise ship A cruise ship is already 200 m from the dock, moving away at a constant 5 m s⁻¹. A smaller boat starts from the dock at the same moment, accelerating at 0.5 m s⁻² from rest. Find how far from the dock the boat catches the cruise ship. Cruise ship's displacement from the dock: s_ship = 200 + 5t (starts 200 m out, moves at constant 5 m s⁻¹). Boat's displacement from the dock: s_boat = ½(0.5)t² = 0.25t² (starts from rest at the dock, accelerating at 0.5 m s⁻²). They meet when s_boat = s_ship: 0.25t² = 200 + 5t. Rearranging: 0.25t² − 5t − 200 = 0. Solve this quadratic for t (taking the positive root, since negative time has no meaning here). Substitute the positive t back into either displacement equation to find the distance from the dock at which they meet. ✓ Setting the two displacement expressions equal, rather than comparing speeds, is the key step — the boat catches up when its TOTAL distance travelled matches the ship's total distance from the same reference point. ✕

Comparing speeds instead of displacements

A common wrong approach compares the two objects' SPEEDS to decide when they "meet" — but matching speed only tells you when the gap between them stops growing (or starts shrinking), not when they are at the same position. The correct condition for meeting is always EQUAL DISPLACEMENT from the same shared reference point, not equal speed.

When there is no head start at all

  • If both objects start from the SAME point at the SAME time (no offset), the equation simplifies: set the two displacement expressions equal directly, with no constant term to account for.
  • A negative solution for t (before either object started moving) is not physically meaningful — discard it and take the smallest positive solution.
  • If the resulting equation has no real positive solution, the two objects never actually meet within the scenario described — say so explicitly rather than forcing an answer.

Practise just this