determine acceleration using the gradient of a velocity–time graph

Equations of Motion – Determining Acceleration from a \cdot elocity–Time Graph

Learning Objective

By the end of this lesson you should be able to determine the acceleration of an object by finding the gradient (slope) of its velocity–time graph.

Key Concepts

• Velocity ($v$) is the rate of change of displacement with respect to time.
• Acceleration ($a$) is the rate of change of velocity with respect to time.
• On a $v$–$t$ graph, the gradient of a straight‑line segment equals the constant acceleration during that interval.

Relevant Equations of Motion

Equation	When to Use
$$v = u + at$$	Relates final velocity $v$, initial velocity $u$, constant acceleration $a$, and time $t$.
$$s = ut + \frac{1}{2}at^{2}$$	Gives displacement $s$ for constant acceleration.
$$v^{2} = u^{2} + 2as$$	Eliminates time; relates velocities, acceleration and displacement.

Why the Gradient Gives Acceleration

For a straight line on a $v$–$t$ graph the relationship between $v$ and $t$ is linear:

$$v = mt + c$$

Comparing with $v = u + at$, the slope $m$ is the acceleration $a$, and the intercept $c$ is the initial velocity $u$.

Step‑by‑Step Procedure

1. Identify the portion of the $v$–$t$ graph that represents constant acceleration (a straight‑line segment).
2. Choose two points on that segment: $(t_{1}, v_{1})$ and $(t_{2}, v_{2})$.
3. Calculate the gradient: $$a = \frac{\Delta v}{\Delta t} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$$
4. State the direction of the acceleration (positive if the line slopes upwards, negative if downwards).
5. Use the obtained $a$ in other equations of motion if required.

Worked Example

A car accelerates uniformly from rest. The velocity–time graph shows that at $t = 4\ \text{s}$ the velocity is $12\ \text{m s}^{-1}$. Determine the acceleration.

1. Choose points: $(0\ \text{s}, 0\ \text{m s}^{-1})$ and $(4\ \text{s}, 12\ \text{m s}^{-1})$.
2. Apply the gradient formula: $$a = \frac{12\ \text{m s}^{-1} - 0\ \text{m s}^{-1}}{4\ \text{s} - 0\ \text{s}} = \frac{12}{4} = 3\ \text{m s}^{-2}$$
3. Since the line slopes upwards, the acceleration is positive: $a = +3\ \text{m s}^{-2}$.

Common Mistakes to Avoid

• Using curved portions of the graph – only straight‑line sections represent constant acceleration.
• Mixing up units – ensure velocity is in $\text{m s}^{-1}$ and time in seconds.
• Reading the gradient from a graph that is not drawn to scale – always verify the scale on the axes.

Summary

The acceleration of an object moving with constant acceleration can be read directly from the gradient of its velocity–time graph. The formula $a = \Delta v / \Delta t$ provides a quick and reliable method, linking the graphical representation to the algebraic equations of motion.

Practice Questions

1. A cyclist speeds up uniformly from $5\ \text{m s}^{-1}$ to $15\ \text{m s}^{-1}$ in $5\ \text{s}$. Determine the acceleration using the gradient method.
2. On a $v$–$t$ graph, a line falls from $20\ \text{m s}^{-1}$ at $t = 2\ \text{s}$ to $0\ \text{m s}^{-1}$ at $t = 6\ \text{s}$. What is the magnitude and direction of the acceleration?
3. Explain why a horizontal line on a $v$–$t$ graph corresponds to zero acceleration.