Lesson Explainer: Acceleration over Time Physics • 9th Grade

In this explainer, we will learn how to analyze the motion of objects that change their velocity in some amount of time, by using the formula for acceleration, 𝑎=Δ𝑣Δ𝑡.

Acceleration of an object occurs when an object changes velocity.

The following figure shows a velocity–time graph for an object that is initially at rest and that increases in velocity as time passes.

The object is at rest when the value of time shown on the graph is zero. The object begins to move as its velocity increases to become nonzero. At the instant that the value of time is zero, the velocity of the object is also zero. However, only when the value of the time is greater than zero is the velocity of the object nonzero.

We see then that when an object accelerates, the change in its velocity must occur in some time interval.

The acceleration, 𝑎, of an object is related to the change in the velocity of the object, Δ𝑣, and the time interval in which the velocity changes, Δ𝑡, by the formula 𝑎=Δ𝑣Δ𝑡.

Acceleration and velocity are both vector quantities related by time. The quantity Δ𝑡 always increases and so it must be positive.

From this, we see that if 𝑎 is positive, then Δ𝑣 is also positive. Equally, if 𝑎 is negative, then Δ𝑣 is negative. We see then that the direction in which the velocity of the object changes is the direction of the acceleration of the object.

The SI unit of velocity is the metre per second (m/s). The SI unit of time is the second (s). The SI unit of acceleration is therefore given by =×=.mssmssms

Let us look at an example involving an accelerating object.

Example 1: Determining the Velocity and Acceleration of an Accelerating Object

A car accelerates, starting at the time 𝑡=0s. The velocity of the car at 2 s intervals after it starts to accelerate is shown in the diagram.

  • What is the velocity of the car at 𝑡=6s?
  • What is the acceleration of the car?

Answer

We see that the car has an initial velocity of zero, so it is initially at rest.

We also see that for each 2 seconds, the velocity of the car increases by 5 m/s.

The car is stated to accelerate uniformly, so the change in the velocity of the car in each time interval must be the same.

At a time of 𝑡=6s, the number of 2-second intervals in which the car has accelerated is given by 62=3.

The total increase in the velocity of the car must then be 3×5=15/.ms

The initial velocity of the car is zero, so the velocity of the car after 6 seconds of acceleration is 15 m/s.

The direction in which the velocity of the car changes is the direction of the acceleration of the car. This direction is shown in the question, so we need only determine the magnitude of the acceleration.

The acceleration, 𝑎, of an object is related to the change in the velocity of the object, Δ𝑣, and the time interval in which the velocity changes, Δ𝑡, by the formula 𝑎=Δ𝑣Δ𝑡.

Substituting the values given in the question, we see that 𝑎=52=2.5/.ms

Let us now look at an example in which the change in the velocity of an accelerating object is determined.

Example 2: Determining the Velocity Increase of an Accelerating Object

An object accelerates at 5 m/s2 for 0.25 s. How much does its velocity increase?

Answer

The direction in which the velocity of the object changes is the direction of the acceleration of the object. This direction is not stated in the question, which asks only “how much” the velocity changes, meaning the magnitude of the velocity change.

The acceleration, 𝑎, of an object is related to the change in the velocity of the object, Δ𝑣, and the time interval in which the velocity changes, Δ𝑡, by the formula 𝑎=Δ𝑣Δ𝑡.

Multiplying this formula by Δ𝑡, we obtain 𝑎×Δ𝑡=Δ𝑣Δ𝑡×Δ𝑡=Δ𝑣Δ𝑣=𝑎×Δ𝑡.

The increase in the velocity of an accelerating object is the acceleration of the object multiplied by the time for which it accelerates.

Substituting the values given in the question, we see that Δ𝑣=5×0.25=1.25/.ms

An object that falls freely is accelerated only by the gravitational force of the Earth. Near the surface of Earth, the acceleration of such an object is a constant 9.8 m/s2.

Let us look at an example of the acceleration of an object that falls freely.

Example 3: Determining the Velocity Increase of a Freely Falling Object

A skydiver accelerates downward at a rate of 9.8 m/s2. How much does their downward velocity increase in 0.67 seconds? Round your answer to two decimal places.

Answer

The direction in which the velocity of the skydiver changes is the direction of the acceleration of the skydiver, vertically downward.

The acceleration, 𝑎, of an object is related to the change in the velocity of the object, Δ𝑣, and the time interval in which the velocity changes, Δ𝑡, by the formula 𝑎=Δ𝑣Δ𝑡.

Multiplying this formula by Δ𝑡, we obtain 𝑎×Δ𝑡=Δ𝑣Δ𝑡×Δ𝑡=Δ𝑣Δ𝑣=𝑎×Δ𝑡.

Substituting the values given in the question, we see that Δ𝑣=9.8×0.67/.ms

To two decimal places, this is 6.57 m/s2.

Acceleration of an object refers to any change in the velocity of any object, not only objects that are initially at rest. An object can have a nonzero velocity before it accelerates.

The changes in the acceleration of an object can be represented in a velocity–time graph. As acceleration is the time rate of change of velocity, the gradient of a velocity–time graph equals the acceleration of an object.

The following figure shows a velocity–time graph for two objects that increase in velocity.

The object that changes its velocity from 𝑢 to 𝑣 is initially at rest, as 𝑢 is zero. The object that changes its velocity from 𝑢 to 𝑣 is initially moving with a nonzero velocity, 𝑢.

For the graph, it is the case that 𝑣𝑢=𝑣𝑢.

Hence, both objects change their velocity equally. The velocities of both objects change in the same time interval, so the accelerations of the objects are equal.

Let us look at an example involving objects accelerating with initial velocities that are nonzero.

Example 4: Identifying a Velocity–Time Graph from a Description of Motion

Select the velocity–time graph that best matches the following description of motion:

A boat moves at constant speed through the water, then accelerates for a short time, and then continues moving at a higher constant speed.

Answer

The boat is described as initially moving at constant speed. The velocity of the boat must initially have a constant magnitude. On a velocity–time graph, the line of best fit must initially have a constant value and, hence, be horizontal. This is only true in graphs (C) and (E).

The line of best fit in each of graphs (C) and (E) changes from being horizontal to having a positive gradient. In graph (C), the line is straight, and in graph (E), the line is curved. Either a straight line or a curved line could indicate acceleration, as they both represent changes in velocity. The straight line corresponds to a constant value of acceleration and the curved line corresponds to a changing value of acceleration. The question does not say whether the acceleration of the boat is constant.

After the boat accelerates, it again moves at a constant speed. The gradient of the graph must then be horizontal for the last part of the motion of the boat. In graph (C), the line is horizontal, and in graph (E), the line has a constant positive gradient, so it is accelerating.

Only graph (C) shows an object moving at constant speed, accelerating to a greater constant speed, and then moving at that greater constant speed.

We have seen that, just like speed or velocity, acceleration can be constant or can vary.

The following figure shows a velocity–time graph for two objects that change velocity from an initial value of 𝑢 to a final value of 𝑣.

Knowing that the acceleration of an object equals the gradient of the velocity–time graph, we see that the acceleration of the object corresponding to the green line is constant but the acceleration of the object corresponding to the blue line varies.

Both objects have the same average acceleration however, as they both change from having velocity 𝑢 to having velocity 𝑣 in the same time interval.

We can see that if the acceleration of an object varies, then it has an average acceleration that may or may not be equal to its acceleration at a particular instant.

Just as with speed and velocity, the average acceleration of an object with a constant acceleration is equal to that constant acceleration.

Let us now look at an example in which the average acceleration of a moving object is determined.

Example 5: Determining the Average Acceleration of a Moving Object

An airplane flying at a velocity of 245 m/s is hit by a strong tailwind. The gust of wind lasts for 2.7 seconds and the airplane’s velocity afterward is 263 m/s. What is the average acceleration of the airplane by the gust of wind? Round your answer to one decimal place.

Answer

The direction in which the velocity of the object changes is the direction of the acceleration of the object. This direction is the direction in which the airplane flies.

The question asks for the average acceleration of the airplane during the gust of wind. For this example, it makes no difference whether we determine the average acceleration of the airplane or the constant acceleration of the airplane as the average acceleration is determined in the same way as if the airplane had a constant acceleration.

The acceleration, 𝑎, of an object is related to the change in the velocity of the object, Δ𝑣, and the time interval in which the velocity changes, Δ𝑡, by the formula 𝑎=Δ𝑣Δ𝑡.

The initial velocity of the airplane is 245 m/s and the final velocity is 263 m/s. The value of Δ𝑣 is therefore given by Δ𝑣=263245=18/.ms

The time that the gust of wind lasts for is stated to be 2.7 seconds.

Substituting these values, we see that 𝑎=182.7/.ms

To one decimal place, this is 6.7 m/s2.

The acceleration of the object does not need to be in the direction of the velocity of the object that accelerates. Acceleration is defined as the time rate of change of velocity, not the increase in velocity. A decrease in the velocity of an object is as much a change of its velocity as an increase in velocity.

A change in the velocity of an object that results in the velocity after the change being closer to zero than before the change is sometimes called a deceleration. A deceleration is just an acceleration that is in the opposite direction to the initial motion of the object.

Let us look at an example involving a deceleration.

Example 6: Determining the Velocity of a Decelerating Object at Different Times

The velocity of a car at different times is shown in the diagram. The car is accelerating uniformly.

  • Find 𝑣.
  • Find 𝑣.
  • Find 𝑣.

Answer

The diagram shows the velocity of a car at 2-second time intervals between 𝑡=0seconds and 𝑡=10seconds. The diagram does not show the positions of the car at these times.

The car is accelerating uniformly, so the change in the velocity of the car must be the same in each time interval. The velocity of the car is decreasing by 5 m/s in each time interval shown.

The value of 𝑣 is the velocity at 𝑡=4s minus 5 m/s. This is given by 105=5/.ms

This is the velocity of the car at 𝑡=6s.

The value of 𝑣 is the velocity at 𝑡=6s minus 5 m/s. This is given by 55=0/.ms

This is the velocity of the car at 𝑡=8s. It is useful to notice that the car is at rest at this instant.

The value of 𝑣 is the velocity at 𝑡=8s minus 5 m/s. This is given by 05=5/.ms

The velocity at 𝑡=10s is negative. This shows that the direction of the velocity is opposite that of the initial velocity. The car is therefore reversing for values of 𝑡 greater than 8 seconds.

Another Solution:

Since the acceleration is uniform, then it has a constant value across the motion of the car. We can calculate this acceleration using the formula 𝑎=Δ𝑣Δ𝑡=𝑣𝑣𝑡𝑡.

Since the acceleration is constant, we can calculate a by choosing any time interval Δ𝑡 and calculate the change in velocity in this time interval Δ𝑡.

Let us choose the time interval between 𝑡=2s and 𝑡=0; then, 𝑡=2s, 𝑡=0s, 𝑣=15 m/s, and 𝑣=20 m/s.

Therefore, 𝑎=152020=52=2.5/.ms

Note that the acceleration being negative means that the velocity of the car decreases with time.

To calculate the unknown velocities in the question, let us arrange the acceleration formula to be 𝑣𝑣=𝑎(𝑡𝑡).

To find 𝑣, then, 𝑣=𝑣, 𝑣=10/ms, 𝑡=4s, and 𝑡=6s.

Therefore, 𝑣10=2.5×(64)𝑣10=5.

Adding 10 to both sides of the equation, we get 𝑣=5+10.

Therefore, 𝑣=5/ms, which is the same value obtained in the other solution.

Let us look at an example involving interpreting the velocity–time graph of an object that reverses direction.

Example 7: Interpreting the Velocity–Time Graph of an Accelerating Object

Which of the following descriptions best matches the motion plotted in the velocity–time graph shown?

  1. An object accelerates in the opposite direction to its velocity, then moves at a constant speed, then accelerates in the same direction that it accelerated before but at a greater rate, and then moves at a greater constant speed.
  2. An object moves at a constant speed, then stops, then starts moving at a lower constant speed, and then stops again.
  3. An object decelerates, then stops, then decelerates at a greater rate, and then stops again.
  4. An object decelerates, then stops, then decelerates at a greater rate in the opposite direction to the previous acceleration, and then stops again.
  5. An object accelerates in the opposite direction to its velocity, then moves at a constant speed, then accelerates in the opposite direction that it accelerated before and at a lower rate, and then moves at a greater constant speed.

Answer

At the start of the graph, the gradient of the line of best fit is negative. From this, we can immediately eliminate option 2, as the object cannot initially have a constant speed if the magnitude of its velocity is decreasing.

In the part of the graph where the line of best fit is horizontal and above the time axis, the object moves at constant, nonzero speed. From this, we can immediately eliminate options 3 and 4, which state that the object stops after its initial deceleration.

This leaves only options 1 and 5, which are very similar. The only difference between these options concerns the third part of the graph in which the velocity changes. In this part of the graph, option 1 states that the acceleration of the object is in the same direction as its initial acceleration, and option 5 states that the acceleration of the object is in the opposite direction to its initial acceleration.

For both parts of the graph where the velocity changes, the line has a nonzero gradient. In both cases, the gradients are negative. The direction of the acceleration is therefore the same in both cases. Option 1 is correct.

Let us now summarize what has been learned in these examples.

Key Points

  • The acceleration, 𝑎, of an object is related to the change in the velocity of the object, Δ𝑣, and the time interval in which the velocity changes, Δ𝑡, by the formula 𝑎=Δ𝑣Δ𝑡.
  • The SI unit of acceleration is metre per second squared (m/s2).
  • The direction of the acceleration of an object does not have to be the same as that of the initial velocity of the object. The direction of the acceleration is the direction of the change of the velocity of the object.

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