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MYP Physics · topic MYP-02.02

Distance, displacement, speed, and velocity

Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.

Extra challenging questions

Work through the linked stages and explain what each result means.

Extra challenging · 01

The journey that ends closer but travels farther

Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. Take east as positive. A robot's velocity varies linearly between the supplied points. At t = 0 it is 2 m east of the origin. The graph describes one-dimensional motion; negative velocity means westward motion. Supplied data (Time / s; Velocity east / m s⁻¹): (0, 4), (2, 4), (4, 0), (6, -4), (8, -4). (a) Determine displacement, total distance, final position, average speed and average velocity for the full 8 s. [3 marks] (b) Find the acceleration between 2 and 6 s and explain whether the robot is slowing throughout that interval. [3 marks]

Explore the question data

Signed robot velocity

Supplied question data. Read the labelled axes and use the observations to test the physical model.

Vertical axis (y): Velocity east / m s⁻¹

Signed robot velocitySupplied question data. Read the labelled axes and use the observations to test the physical model. Horizontal axis: Time / s. Vertical axis: Velocity east / m s⁻¹. Exact values are available in the data table below.-4-202402468xy

Horizontal axis (x): Time / s

Reading 1Time / s: 0Velocity east / m s⁻¹: 4
Read the exact data as a table
Signed robot velocity — supplied values
Time / sVelocity east / m s⁻¹
04
24
40
6-4
8-4

These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.

Show a hint

Begin with part (a): Determine displacement, total distance, final position, average speed and average velocity for the full 8 s.

Reveal the worked solution
  1. Signed graph areas are +8, +4, −4 and −8 m, so their sum is zero displacement.
  2. Distance adds the magnitudes, giving 24 m. The final position is the initial 2 m plus zero displacement.
  3. Average speed = 24/8 = 3 m s⁻¹, whereas average velocity = 0/8 = 0. Distance and speed are scalars; displacement and velocity require direction.
  4. The gradient is (−4−4)/(6−2) = −2 m s⁻² throughout the interval.
  5. Before 4 s acceleration opposes velocity, so speed decreases. After 4 s acceleration and velocity both point west, so speed increases.
  6. A negative acceleration does not alone mean slowing. Compare its direction with the instantaneous velocity.

Answer (a) Displacement = 0 m; distance = 24 m; final position = 2 m east; average speed = 3 m s⁻¹; average velocity = 0. (b) Acceleration = −2 m s⁻²; it slows until 4 s and then speeds up westward.

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