## 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?

**Show/Hide Answers**

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.

**Show/Hide Answers**

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?

**Show/Hide Answers**

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?

**Show/Hide Answers**

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?

**Show/Hide Answers**

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.

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.

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A very informative site .

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!

area under graph for the acceleration time graph gives u the ‘Change’ in velocity. when an object is stationary , it has zero velocity. Stationary objects wont have any acceleration either as they r not moving thus don,t have any velocity and acceleration basically is the rate of change of velocity. hope this helped

I would like to know the answer please.

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