Supercharge your learning!

Use adaptive quiz-based learning to study this topic faster and more effectively.

Kinematics graphs

A distance-time graph shows the time $$t$$ on the $$x$$-axis and the distance an object has travelled $$d$$ on the $$y$$-axis.

The gradient of the graph is equal to the speed of the object at time $$t$$.

A straight line on a distance-time graph indicates uniform speed. When the straight line is horizontal, the speed is zero. In other words, the object is at rest.

A curved line on a distance-time graph indicates motion at non-uniform speed. A line curving upwards indicates that the object is accelerating. A line with a gradient that is decreasing indicates that the object is decelerating.

In section A the object is accelerating. In section B the object is at rest. In section C the object is travelling at uniform speed.

A speed-time graph shows the time on the x-axis and speed on the y-axis.

The gradient of the graph gives the acceleration.

  • A horizontal line indicates that the object is moving at a constant velocity.
  • A positive gradient indicates that the object is accelerating.
  • A negative gradient indicates that the object is decelerating.
  • A curved line indicates non-uniform acceleration.
 The graph shows a vehicle accelerating and then driving at a constant speed before decelerating until stopped

The area under the graph indicates the distance travelled.

To work out the distance travelled during $$\Tblue{\text{section A}}$$ on the graph we need to find the area of the triangle underneath the graph. $$$\Tblue{\text{distance}} = \frac{1}{2} \times 10 \us \times 20 \umps = 100 \um$$$

The distance travelled during $$\Tred{\text{section B}}$$ is equal to the area of the rectangle under the graph. $$$\Tred{\text{distance}} = 10 \us \times 20 \umps = 200 \um$$$

The total distance travelled is equal to the sum of the three sections $$$\text{total distance} = 100 \um + 200 \um + 200 \um = 500 \um.$$$

A displacement-time graph shows the time $$t$$ on the $$x$$-axis and the displacement $$\vecphy{s}$$ of an object on the $$y$$-axis. The object in question is assumed to undergo linear motion.

Negative $$\vecphy{s}$$ values imply that the object has moved past its starting point in the opposite direction.

The gradient of the graph is equal to the velocity at time $$t$$.

A straight line in the displacement-time graph indicates constant velocity. When the straight line is horizontal, $$v=0 \text{ m/s}$$. In other words, the object stands still.

  • A - object travels at uniform speed of $$2 \umps$$.
  • B - object is at rest $$2 \um$$ from origin.
  • C - object travels at $$-2 \umps$$, passing origin on the way.
  • D - object is at rest $$-2 \um$$ from origin.
  • E - object travels at uniform speed of $$2 \umps$$, returning to origin.

The velocity-time graph shows the time $$t$$ on the x-axis and velocity $$\vecphy{v}$$ on the y-axis. Similar to the displacement-time graph, the object is assumed to undergo linear motion with a defined positive direction.

Negative $$v$$ values would imply that the object is moving in the opposite direction.

The gradient of the graph gives the acceleration. Therefore, a horizontal line indicates that the object is moving at a constant velocity.

A negative gradient indicates that the velocity in the positive direction is decreasing.

The area between the graph and the line $$v=0$$ indicates the magnitude of displacement during the period.

If the graph crosses the t-axis and the area above is equal to the area below, the object has returned to its origin.

 The graph shows a vehicle accelerating and then driving at a steady pace before stopping at the destination and then returning to the point of origin.

The acceleration-time graph shows the time $$t$$ on the x-axis and acceleration $$\vecphy{a}$$ on the y-axis. As with displacement-time and velocity-time graphs, the object is assumed to undergo linear motion.

The area under the graph gives the increase in velocity between two particular points in time. If the movement started from a stationary position, the sign of the total area indicates the direction of the movement (forward or backward).

The overall increase in velocity is equivalent to the area under the graph.