# 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**.

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 **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 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).