Calculate acceleration from the gradient of a speed-time graph

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – Motion: Calculating Acceleration from a Speed‑Time Graph

1.2 Motion – Calculating Acceleration from the Gradient of a Speed‑Time Graph

Learning Objective

By the end of this lesson you should be able to calculate the acceleration of an object by determining the gradient (slope) of its speed‑time graph.

Key Concepts

  • Acceleration is the rate of change of speed with time.

  • Mathematically, \$a = \dfrac{\Delta v}{\Delta t}\$ where \$\Delta v\$ is the change in speed and \$\Delta t\$ is the change in time.

  • On a speed‑time graph, the gradient (rise over run) represents acceleration.

Understanding the Gradient

For a straight‑line segment on a speed‑time graph:

\$\text{Gradient} = \frac{\text{vertical change (Δv)}}{\text{horizontal change (Δt)}} = \frac{\Delta v}{\Delta t} = a\$

If the line is horizontal (gradient = 0), the acceleration is zero and the object moves at constant speed.

Suggested diagram: A speed‑time graph showing a straight‑line segment with marked Δv and Δt.

Step‑by‑Step Procedure

StepActionWhat to Record
1Identify the straight‑line portion of the graph whose acceleration you need.Start and end points (time and speed).
2Read the speed values at the two points (Δv).Δv = \$v{final} - v{initial}\$ (in m s⁻¹).
3Read the corresponding times (Δt).Δt = \$t{final} - t{initial}\$ (in s).
4Calculate the gradient using \$a = \dfrac{\Delta v}{\Delta t}\$.Acceleration in m s⁻².
5State the direction of acceleration (positive if speed is increasing, negative if decreasing).Sign of \$a\$.

Worked Example

Consider a car whose speed‑time graph shows a straight line from (2 s, 4 m s⁻¹) to (6 s, 12 m s⁻¹).

  1. Δv = \$12 - 4 = 8\ \text{m s}^{-1}\$

  2. Δt = \$6 - 2 = 4\ \text{s}\$

  3. Acceleration \$a = \dfrac{8}{4} = 2\ \text{m s}^{-2}\$

  4. The line slopes upwards, so the acceleration is positive (speed is increasing).

Common Mistakes to Avoid

  • Using the total time of the experiment instead of the time interval for the chosen segment.

  • Reading speed values from the wrong axis (e.g., reading distance instead of speed).

  • Ignoring the sign of the gradient; a downward slope indicates negative acceleration (deceleration).

  • Mixing units – ensure speed is in m s⁻¹ and time in seconds.

Practice Questions

  1. From a speed‑time graph, a cyclist’s speed increases from 3 m s⁻¹ at \$t = 5\ \text{s}\$ to 11 m s⁻¹ at \$t = 9\ \text{s}\$. Calculate the acceleration.

  2. A ball rolls down a slope and its speed‑time graph shows a straight line from (0 s, 0 m s⁻¹) to (4 s, 8 m s⁻¹). What is the acceleration? State whether the ball is speeding up or slowing down.

  3. On a speed‑time graph a car moves at a constant speed of 15 m s⁻¹ for 10 s, then its speed decreases uniformly to 5 m s⁻¹ over the next 5 s. Determine the magnitude and direction of the acceleration during the deceleration phase.

Summary

The gradient of a straight‑line segment on a speed‑time graph gives the acceleration of the object during that interval. By measuring the change in speed (Δv) and the corresponding change in time (Δt) and applying \$a = \dfrac{\Delta v}{\Delta t}\$, you can obtain the acceleration in m s⁻². Remember to keep track of units and the sign of the gradient.