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Multi-stage motion | IB Physics Practice | Physics Insight
IB Physics/A.1 Kinematics/Question

Question pattern: Multi-stage motion

Original question 1 of 1 in this pattern

HardCalculate & apply6 marks8 minutes

Multi-stage motion

A cyclist passes point P at 5.0 m s⁻¹. The cyclist then accelerates uniformly at 1.2 m s⁻² for 6.0 s, travels at the resulting constant speed for 4.0 s, and finally decelerates uniformly at 2.0 m s⁻² until coming to rest.

Calculate:

a) the maximum speed;

b) the total distance travelled from P until the cyclist stops;

c) the average speed for the complete journey.

Worked answer

Solution

  1. Step 1

    Maximum speed: v = 5.0 + 1.2 × 6.0 = 12.2 m s⁻¹.

  2. Step 2

    First-stage distance: s₁ = 5.0(6.0) + 1/2(1.2)(6.0²) = 51.6 m.

  3. Step 3

    Constant-speed distance: s₂ = 12.2 × 4.0 = 48.8 m.

  4. Step 4

    Deceleration time: t₃ = 12.2 / 2.0 = 6.10 s.

  5. Step 5

    Final-stage distance: s₃ = ((12.2 + 0) / 2) × 6.10 = 37.2 m.

  6. Step 6

    Total distance: 137.6 m.

  7. Step 7

    Total time: 16.1 s.

  8. Step 8

    Average speed: 137.6 / 16.1 = 8.55 m s⁻¹.

Final answer

Maximum speed = 12.2 m s⁻¹; total distance ≈ 138 m; average speed = 8.55 m s⁻¹.

Common mistake

Using the mean of the initial and final velocities as the average speed for the entire multi-stage journey.

Concepts tested

  • multi-stage motion
  • acceleration
  • velocity-time reasoning

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