Show/Hide Sub-topics (Speed, Velocity and Acceleration | O Level)
Show/Hide Sub-topics (Kinematics | A Level)

Distance-Time Graphs

• For a distance-time graph, the distance never decreases.
• When the object is stationary, the distance-time graph will be horizontal.
• The gradient of a distance-time graph is the instantaneous speed of the object.
• For straight line with positive gradient, it means that the object is travelling at uniform speed
• There is no straight line with negative gradient (as the distance never decreases)
• For curves, it means that the object is travelling at non-uniform speed

Displacement-Time Graphs

• The details are similar as distance-time graphs, except that the distance is now displacement, and speed is now velocity.
• One minor difference: There is a straight line with negative gradient, it means that the object is travelling at uniform velocity in the opposite direction.

Velocity-Time Graphs

• When the object is stationary, it is a straight horizontal line at 0.
• When the object is undergoing uniform motion, it is a straight horizontal line at $v \, \text{m s}^{-1}$, where v is the velocity of the object.
• For straight line with positive gradient, it means that the object is accelerating.
• For straight line with negative gradient, it means that the object is decelerating.
• For curves, it means that the acceleration of the object is changing.
• The area under the graph is the change in the displacement of the object.

Acceleration-Time Graphs

• Area under graph is the change in the velocity of the object

Summary of Kinematics Graphs

The figure below shows the displacement-time graph, velocity-time graph and acceleration-time graph for the respective state of motion. It serves as a summary of the text above.

The figure below shows the relationship between displacement-time graph, velocity-time graph and acceleration-time graph.

Self-Test Questions

Can you tell from a displacement-time graph whether an object is stationary?

Yes. If the object is stationary, it will appear as a horizontal line on a displacement-time graph.

How can you obtain the average velocity and instantaneous velocity from a displacement-time graph.

The average velocity can be found by using $\frac{\text{total displacement}}{\text{total time taken}}$.

The instantaneous velocity at a point in time can be found from the gradient of the tangent to that point in time.

Can you tell from a velocity-time graph whether an object is stationary?

Yes. If the object is stationary, the velocity-time graph will be a horizontal line at $\text{v}=0$.

How would you obtain the acceleration of an object from a velocity-time graph? What does the area under a velocity-time graph represent?

The acceleration of an object at a point in time can be obtained from the gradient of the tangent to that point in time.

The area under a velocity-time graph represents the total distance traveled.

Can you tell from an acceleration-time graph whether an object is stationary?

No, you cannot. Do you know why?

Hint: Refer to the summary of Kinematics graphs (located above).

Drop a comment below if you cannot figure out the answer.

21 thoughts on “Reading Kinematics Graphs”

1. For those finding the answe, look below: stationary object means 0 velocity hence velocity donot changes so acc=0 (acc is the rate at which velocity changes). Now acceleration is also 0 when object moves with constant velocity.. as a result we cannot predict whether or not object is stationary from acceleration time graph.

• The answer is that an object is stationary when it’s velocity is zero (so the acceleration will also be zero as acceleration is the rate of change of velocity) or also when it’s velocity is constant. Therefore, we can’t deduce from an acceleration time graph that whether an object is stationary or not.

2. A very useful website. Helped me a lot while preparing my exam

3. A very informative site .

4. Can you tell from an acceleration-time graph whether an object is stationary?

No, you cannot. Do you know why?

Drop a comment below if you cannot figure out the answer.
I would like to know the answer!