AshPhys

Deducing displacement

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.5Sections 2.5

Goal: Find displacement from the area under a velocity–time graph.

01The big picture

The area between a velocity–time graph and the time axis is the displacement: velocity × time, added up over the whole journey. Split the area into rectangles and triangles and add them.

Area below the time axis is negative displacement (moving back). For uniform acceleration the area is a trapezium, which gives s=12(u+v)ts = \tfrac{1}{2}(u + v)t.

Area under a v–t graph [paraphrase]
The displacement during that time interval.
Fig. 1 · Area under a v–t graph
Area under a v–t graphVelocity against time. The velocity rises uniformly from 0 to 10 m/s in 4 s, stays at 10 m/s until 10 s, then falls to 0 at 12 s. The area under the line is shaded and split by dashed lines at 4 s and 10 s into regions 1, 2 and 3.v / m s⁻¹t / s10041012123
The shaded area is the displacement: two triangles and a rectangle.

02Key equations

Displacement = area under v–t
s=area under the v–t graphs = \text{area under the } v\text{–}t \text{ graph}
Uniform acceleration (trapezium)
s=12(u+v) ts = \tfrac{1}{2}(u + v)\,t
Symbols and units
  • ssdisplacementm\mathrm{m}
  • u, vu,\ vinitial, final velocitym s−1\mathrm{m\,s^{-1}}
  • tttimes\mathrm{s}

03Sample problems with solutions

Sample 1 · Adding up the areas
Use the graph in Fig. 1 to find the total displacement and the average velocity for the 12 s journey.

  1. Area 1 (triangle) + area 2 (rectangle) + area 3 (triangle):

    s=12(4)(10)+(6)(10)+12(2)(10)=20+60+10=90 ms = \tfrac{1}{2}(4)(10) + (6)(10) + \tfrac{1}{2}(2)(10) = 20 + 60 + 10 = 90\ \mathrm{m}

  2. Average velocity = total displacement ÷ total time:

    vˉ=9012=7.5 m s−1\bar v = \frac{90}{12} = 7.5\ \mathrm{m\,s^{-1}}
Answer: 90 m; average velocity 7.5 m s⁻¹
Unit check: m s−1×s=m ✓\mathrm{m\,s^{-1}} \times \mathrm{s} = \mathrm{m}\ \checkmark

04Practice problems

  1. P1
    A car travels at a constant 8.0 m s⁻¹ for 15 s. Find the displacement.
    Final answer120 m
    Answer: 120 m
  2. P2
    A car accelerates uniformly from rest to 20 m s⁻¹ in 10 s. Find the distance travelled.
    Final answer100 m
    Answer: 100 m
  3. P3
    A cyclist accelerates uniformly from 6.0 m s⁻¹ to 14 m s⁻¹ in 5.0 s. Find the displacement.
    Final answer50 m
    Answer: 50 m
  4. P4
    Part of a v–t graph lies below the time axis. What does that part show?
    Final answerMoving in the negative direction (negative displacement)
    Answer: Moving in the negative direction (negative displacement)

05Other useful information

Common mistakes

  • ✗ Wrong: Use s=vts = vt when the velocity is changing.

    ✓ Right: Use the area: for uniform acceleration s=12(u+v)ts = \tfrac{1}{2}(u + v)t.

Exam tips

  • Label the shapes 1, 2, 3 on the graph and write each area with its unit.

What the syllabus asks

  • 9702 topic 2: displacement from the area under a v–t graph. Coursebook 2.5.

Explore: Practice questions for this lesson