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1.3.2Forces and Their Effects

Velocity-Time Graphs

5 min read 5 parts Lesson 13 of 32 1 quiz

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Section 1 of 5Reading the Shape

1.3.2 Velocity-Time Graphs

On a velocity-time graph the vertical axis shows velocity and the horizontal axis shows time.

Sloping line
  • A straight line with a gradient shows constant acceleration, so the velocity changes by equal amounts every second.
Horizontal line above zero
  • Constant velocity, so the acceleration is zero and the object keeps the same speed in a straight line.
Horizontal line at zero
  • The velocity is zero, so the object is stationary.
Curved line
  • The acceleration is changing, and the gradient at a point gives the acceleration at that instant.

Section 2 of 5Calculating Acceleration from the Gradient

The gradient is found from a large triangle drawn directly on the graph.

1
Choose a straight section of the line and draw a large gradient triangle on the graph itself.
2
Read Δv as the vertical change in velocity and Δt as the horizontal change in time.
3
Divide Δv by Δt to obtain the acceleration in m/s².
4
Use the whole straight section where possible, because a longer line reduces the effect of reading errors.

Section 3 of 5Calculating Distance from the Area

Calculating Distance from the Area
Triangle
  • The area of a triangle is half the base times the height, and it is used when the velocity changes uniformly.
Rectangle
  • The area of a rectangle is the base times the height, and it is used when the velocity is constant.
Total
  • Add the areas of all the sections between the line and the time axis to find the total distance.
Below the axis
  • An area below the time axis represents motion in the opposite direction, so it counts as negative displacement.

Section 4 of 5Worked Example

Question

  • A car accelerates uniformly from rest to 20 m/s in 10 seconds, then travels at 20 m/s for a further 30 seconds.
  • Calculate the acceleration during the first 10 seconds and the total distance travelled.

  1. Acceleration in the first sectiona = Δv ÷ Δt = (20 − 0) ÷ 10 = 2 m/s².
    What the symbols mean: a = acceleration (m s−2) · v = final velocity · u = initial velocity (m s−1) · Δv = change in velocity · t = time taken (s)
  2. Area of the first section, a triangle: ½ × 10 × 20 = 100 m.
  3. Area of the second section, a rectangle: 30 × 20 = 600 m.
  4. Total distance = 100 + 600 = 700 m.

Answer The acceleration is 2 m/s² and the total distance travelled is 700 m.

Section 5 of 5Exam Checklist & Tips

1.3.2 – Key Points for the Exam

  • Velocity-time graph A graph showing how the velocity of an object changes with time.
  • Gradient The change in velocity divided by the change in time, which equals the acceleration.
  • Constant velocity A velocity that does not change, shown by a horizontal line.
  • Area under the graph The region between the line and the time axis, which gives the distance travelled.
  • Uniform acceleration Acceleration that stays the same throughout, shown by a straight sloping line.

Common Mistakes

  • Using the height of the graph as the acceleration, instead of the gradient.
  • Reading a horizontal line as acceleration; a horizontal line on this graph means zero acceleration.
  • Finding a total area from a complex graph without splitting it into separate simple shapes.
  • Applying distance-time rules, such as reading the gradient as speed, to a velocity-time graph.
Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

Constant Acceleration Equations (SUVAT) & Slopes

v = u + at  |  s = ut + ½at2  |  v2 = u2 + 2as  |  s = ½(u + v)t

  • Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
  • Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
  • Free Fall in vacuum: a = g ≈ 9.81 m/s2 downwards (air resistance negligible).