All subjects · Mathematics · WME01 (M1)

Multi-stage motion and two moving objects

Many M1 questions split a journey into stages, or have two objects moving at once. The final velocity of one stage is the initial velocity of the next.

Stages

Treat each stage with its own suvat list. Link them: v at the end of stage 1 = u for stage 2. A velocity-time sketch helps you see the stages.

Two objects

Use the same t for both (or t and t - k if one starts k seconds later). Write each displacement from the same starting point in terms of t, then set them equal (catching up or meeting) or use their sum (moving towards each other).

Checking

Make sure the time you find is in the stage you assumed. For example, if a car stops accelerating at t = 10, a meeting time of 14 means you need the constant speed equation for the last 4 s.

Worked example

A car starts from rest, accelerates at 2 m/s2 for 5 s, travels at constant speed for 10 s, then decelerates at 2.5 m/s2 to rest. Find the total distance.

  1. Stage 1: v = 10 m/s, s = (1/2)(2)(25) = 25 m
  2. Stage 2: s = 10 × 10 = 100 m
  3. Stage 3: 0 = 100 - 5s, s = 20 m

Answer: Total = 145 m

Key idea: End velocity of one stage = start velocity of the next. Two objects: same clock, same origin, then equate positions.

Check yourself

  1. A train starts from rest, accelerates at 0.5 m/s2 for 40 s, then travels at constant speed for 2 minutes. Find the total distance travelled.
    Show answer

    2800 m. Stage 1: v = 20 m/s, s = (1/2)(0.5)(1600) = 400 m. Stage 2: 20 × 120 = 2400 m. Total 2800 m.

  2. Car A passes a stationary car B at a constant 15 m/s. At that instant B starts from rest with acceleration 2.5 m/s2 in the same direction. Find the time taken for B to catch A.
    Show answer

    12 s. 15t = 1.25t2, so t = 12 s (t = 0 is the start).

  3. For the same cars, find the distance B has travelled when it catches A.
    Show answer

    180 m. 15 × 12 = 180 m.

  4. Particle P is projected upwards from the ground at 20 m/s. At the same instant Q is released from rest 40 m vertically above P. Find the time when they meet.
    Show answer

    2 s. Heights: P is 20t - 4.9t2 and Q is 40 - 4.9t2. Equal when 20t = 40, so t = 2 s.

  5. For the same particles P and Q, find the height at which they meet, to 3 significant figures.
    Show answer

    20.4 m. 40 - 4.9(2)2 = 20.4 m.

  6. A cyclist moving at 12 m/s brakes with constant deceleration and stops after 30 m. Find the deceleration.
    Show answer

    2.4 m/s2. 0 = 144 - 60a gives a = 2.4 m/s2.

  7. In a two-stage journey, which quantity carries over from stage 1 to stage 2?
    • The final velocity of stage 1 becomes the initial velocity of stage 2
    • The acceleration of stage 1 is used in stage 2
    • The time of stage 1 equals the time of stage 2
    • The displacement of stage 1 is u for stage 2
    Show answer

    The final velocity of stage 1 becomes the initial velocity of stage 2. The velocity is continuous: the object does not jump in speed between stages.

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Last updated 7 November 2026. Spotted a mistake? Tell us.