Lesson Explainer: Acceleration Science

In this explainer, we will learn how to determine the accelerations of objects that change the speed at which they move.

Recall that if an object moves at constant speed, the value of its speed does not change. An object moving at constant speed covers equal distances in equal time intervals, like the car in the diagram below.

The car above is covering a distance of 20 m every second; in other words, it is moving at a constant speed of 20 m/s.

We commonly encounter situations in which an object’s speed is not constant. For example, the car above might start at rest and gradually increase its speed, as shown in the following diagram.

This time, the distance moved by the car in each 1 s interval changes with each time interval. This means the car’s speed is changing, or the car is accelerating.

Any object whose speed is changing with time is accelerating, and an object’s acceleration is the amount by which its speed is changing. A large acceleration means that the object is increasing its speed more in each 1 s interval.

For example, consider the situations below, representing the motion of three different cars. The three diagrams are shown on a same-distance scale, and the cars’ positions are shown at 1 s intervals.

The purple car in the top row covers the same distance in each second. This means its speed is constant, so its acceleration is zero.

The other two cars cover a greater distance in each subsequent 1 s interval, so they are both accelerating. If we look at the amount by which their speed is increasing, we can see that the gap between the blue cars on the lower row increases by the greatest amount in each interval. The blue car, therefore, has the greatest acceleration.

Example 1: Understanding Acceleration

If the speed of a moving object does not change, which of the following must the acceleration of the object be?

  1. Equal to the speed of the object
  2. Zero

Answer

This example is asking about the definition of acceleration, so we need to recall that acceleration is the amount by which an object’s speed changes. In this example, we have a moving object whose speed does not change. If the speed is not changing, the object has no acceleration. In other words, the acceleration is zero. The correct answer is, therefore, option B: zero.

We have seen that acceleration is the amount by which an object’s speed changes in each second. We can express this in the form of an equation as 𝑎=Δ𝑣Δ𝑡, where 𝑎 is the acceleration, Δ𝑣 is the change in speed, and Δ𝑡 is the change in time. The Δ symbol is the Greek letter delta, which is often used to indicate the change in some quantity.

Consider the car below, which starts from rest at a time of 𝑡=0s and is recorded moving with a speed of 6 m/s at a time of 𝑡=2s.

We can calculate the acceleration of this car using the expression above. We need to know the change in speed, Δ𝑣, over the time interval—that is, the final speed minus the initial speed. We can see from the diagram that the final speed was 6 m/s and the initial speed was 0 m/s, so we can write the change in speed as Δ𝑣=6/0/=6/msmsms.

We also need to know the change in time, Δ𝑡. The time that the second measurement was taken is 2 s, and the initial time was 0 s. We can therefore write Δ𝑡=20=2sss.

With these quantities, we can write the acceleration, 𝑎, mathematically as 𝑎=Δ𝑣Δ𝑡=6/2=3/.mssms

So, the car has acceleration of 3 m/s2.

Notice that the SI base units of acceleration are metres per second squared. This is because we have taken one quantity, speed, with units of metres per second and divided by another quantity, time, with units of seconds. This gives us a unit of metres per second per second, which we can also say as metres per second squared or write as m/s2.

When we worked out the values of the change in speed, Δ𝑣, and the change in time, Δ𝑡, we took the final value minus the initial value of both speed and time. This means we can rewrite the equation for acceleration to include this step, which gives 𝑎=𝑣𝑣𝑡𝑡, where 𝑎 is again the acceleration, 𝑣 is the initial speed, 𝑣 is the final speed, 𝑡 is the initial time, and 𝑡 is the final time.

We can repeat the calculation of the acceleration of the car above using this version of the equation. Now, we have the initial speed 𝑣=0/ms at the initial time 𝑡=0s and the final speed 𝑣=6/ms at the final time 𝑡=2s. Substituting these values in to find the acceleration, we have 𝑎=𝑣𝑣𝑡𝑡=6/0/20=6/2=3/.msmsssmssms

As shown above, these two forms of the equation for acceleration are equivalent to each other, because Δ𝑣=𝑣𝑣 and Δ𝑡=𝑡𝑡. We can use whichever form is most convenient for the quantities given. In situations like this example where we are given the starting and finishing times and speeds, the second version of the equation allows us to substitute the values in directly with no prior calculations and so is more straightforward.

However, we might also encounter situations in which we know the change in quantities, such as the car in the diagram below.

Here, we have a car whose speed has changed by 20 m/s over a time interval of 5 s. We are not told what the start and end speeds are—the car could have started from rest, or it could already have been moving at the start of this time interval. In this case, we do not have enough information to calculate the speed of the car, but we can still determine its acceleration. In this case, we are given Δ𝑣 and Δ𝑡 directly, so we can use the first form of the equation for acceleration to find 𝑎=Δ𝑣Δ𝑡=20/5=4/.mssms

So, the car accelerated by 4 m/s2 over this time interval.

Equation: Acceleration

For an object whose speed changes by an amount Δ𝑣 over a time interval Δ𝑡, its acceleration is given by 𝑎=Δ𝑣Δ𝑡.

If the object started at time 𝑡 with a speed of 𝑣 and finished at time 𝑡 with a speed of 𝑣, we can write this as 𝑎=𝑣𝑣𝑡𝑡.

In the next example, we will look at changes in acceleration.

Example 2: Understanding Uniform Acceleration

If the speed of an object increases the same amount each second, what will happen to the acceleration of the object?

  1. It will not change.
  2. It will be zero.
  3. It will decrease.
  4. It will increase.

Answer

In this example, we are considering how the acceleration of an object changes over time. We are told that an object’s speed increases by the same amount every second, and we need to determine what kind of acceleration the object has.

We are told that the object’s speed is increasing, which means it has some acceleration. We can therefore immediately rule out option B, as the acceleration is not zero.

Recall that acceleration is defined as the change in speed per second. We can write this mathematically as 𝑎=Δ𝑣Δ𝑡, where Δ𝑣 is the change in speed and Δ𝑡 is the change in time. Let’s imagine that we are recording the object’s speed over successive time intervals of 1 s each. This means that whichever time interval we choose, Δ𝑡 is always 1 s.

We are also told that the object’s speed increases by the same amount each second. So, although we do not know the actual value of the speed, we know that the change in speed, Δ𝑣, is the same over every time interval of 1 s.

So, in each time interval, we have the same values for both Δ𝑣 and Δ𝑡. This means that if we take 𝑎=Δ𝑣Δ𝑡 for any time interval of 1 s, we will always get the same answer for 𝑎. The answer is therefore option A: the acceleration of the object will not change.

We can consider the other available options too. If the acceleration were decreasing, we would expect to see the speed increase by a smaller amount in each second. And if the acceleration were increasing, the speed would increase by a greater amount each second.

In the example above, we considered the motion of an object whose acceleration does not change. If the acceleration does not change, it is said to be uniform.

Consider the motion of the car below, whose speed is recorded at 1 s intervals.

Let’s start by considering the first interval. Over the first second, the car started from rest and accelerated to 3 m/s. We can calculate the acceleration as 𝑎=𝑣𝑣𝑡𝑡=3/0/10=3/1=3/.msmsssmssms

So, the acceleration over the first interval is 3 m/s2.

Moving on to the second time interval, the car now starts at a speed of 3 m/s and accelerates to 6 m/s between the times of 1 s and 2 s. Substituting these values in to find the acceleration, we have 𝑎=𝑣𝑣𝑡𝑡=6/3/21=3/1=3/.msmsssmssms

So, the acceleration over the second interval is also 3 m/s2.

We can also carry out the same calculation for the third interval, between the times of 2 s and 3 s, when the car’s speed increases from 6 m/s to 9 m/s. The acceleration is 𝑎=𝑣𝑣𝑡𝑡=9/6/32=3/1=3/,msmsssmssms so we have the same acceleration in the third interval.

As we have seen, it does not matter which interval we choose. When an object is accelerating uniformly, the acceleration is the same over any time interval. This would also be true if we considered only the first and last values. In this case, the car starts from rest at time 𝑡=0s, and we record a speed of 9 m/s at 𝑡=3s. So, the acceleration is 𝑎=𝑣𝑣𝑡𝑡=9/0/30=9/3=3/.msmsssmssms

We could choose any combination of starting and finishing times, and we would always find an acceleration of 3 m/s2. If we know that an object’s acceleration is uniform, we could choose any one interval to calculate the acceleration and this value would hold throughout the object’s motion.

We often encounter situations in which we do not know the car’s speed at any given point, but we know its position or the distance it has moved. Let’s look at the car below, accelerating from rest, whose position is measured at 1 s intervals.

In this case, it is the car’s position, not its speed, that is recorded every second. To calculate the acceleration, we first need to know the speed of the car at each position.

We are given the distance moved by the car in each interval, as well as the time of each interval (1 s). Recall that averagespeeddistancetime=. In the first interval, during which the car moved a distance of 3 m in 1 s, it therefore had an average speed of 31=3/msms.

In the second interval, the car moved a distance of 6 m in 1 s, so it had an average speed of 61=6/msms. Between the first and second intervals, the car’s acceleration was therefore 𝑎=𝑣𝑣𝑡𝑡=6/3/1=3/1=3/.msmssmssms

In the final 1 s interval, the car moved a distance of 9 m. Its average speed over that interval is 91=9/msms. We can calculate the acceleration as 𝑎=𝑣𝑣𝑡𝑡=9/6/1=3/1=3/.msmssmssms

So the acceleration over each interval was the same, 3 m/s, and the car is therefore undergoing uniform acceleration.

In this example, notice that because the positions were recorded at regular time intervals, we were always dividing the distance by the same time value, in this case 1 s, to find the speed. This means that if the distance increases at a constant rate, in this case by an additional 3 m in every second, then the speed also increases at the same rate, in this case 3 m/s, and if the speed increases at a constant rate, the acceleration is uniform.

This means that if we are only given the position of an object at regular time intervals and the distance moved in each time interval increases at a constant rate, we can identify the object as having uniform acceleration.

The next few examples give some more practice identifying and calculating uniform acceleration.

Example 3: Identifying Objects with Uniform Acceleration from Their Positions

The following images show the positions of objects at times one second apart. The objects start from rest. Which image shows an object that has uniform acceleration?

Answer

This example requires us to identify uniform acceleration from an object’s position at different times.

We are told that the positions are recorded at times one second apart. This means that the average speed in each interval increases by the same amount as the distance moved. In order for the acceleration to be uniform, we need the speed to increase by the same amount in each time interval. This means the additional distance moved must also increase by the same amount.

All three of the objects move the same distance in the first second, equal to the width of one of the squares. Looking at option A, we can see that the distance between the second and third blocks is larger than the distance between the first and the second, and the distance between the third and the fourth is larger still. So far, this is what we would expect to see for an object with uniform acceleration. However, the distance between the fourth and fifth blocks is smaller than that between the third and the fourth. This means the speed is no longer increasing but has decreased. This object does not have uniform acceleration, so we can exclude option A.

Looking at option B, we can see that the gap between the second and third blocks is larger than that between the first and the second, and the gap between the third and the fourth is larger still. This is what we would expect to see in the case of uniform acceleration, so B could be the correct answer.

Moving on to option C, we have a larger gap between the second and third blocks than between the first and second, but then the gap between the third and the fourth is the same as that between the second and the third. This means the object has not accelerated in that interval, so option C cannot represent uniform acceleration.

The correct answer is therefore option B.

Example 4: Calculating Acceleration from Change in Speed and Time

A car starts accelerating uniformly from rest. After accelerating for 3 seconds, the car has a speed of 18 metres per second. What is the acceleration of the car?

Answer

In this example, we are given the initial and final speeds of a car and the change in time, and we need to determine the acceleration.

Recall that acceleration, 𝑎, can be found from the equation 𝑎=Δ𝑣Δ𝑡, where Δ𝑣 is the change in speed and Δ𝑡 is the change in time.

Substituting in the values Δ𝑣=18/ms and Δ𝑡=3s, we have 𝑎=Δ𝑣Δ𝑡=18/3=6/.mssms

So, the acceleration of the car is 6 m/s2.

So far, we have only considered objects whose speed is increasing. However, acceleration is any change in speed, even if the speed is decreasing.

In everyday language, we would say that an object that is slowing down is decelerating. In physics, though, this is the same thing as acceleration but in the opposite direction. We therefore say that an object that is slowing down is accelerating, and we use a negative sign to indicate that the speed is decreasing.

For example, consider the car below, whose speed is recorded at 1 s intervals as it slows down.

To calculate the acceleration of this car, we use the usual equation 𝑎=Δ𝑣Δ𝑡=𝑣𝑣𝑡𝑡, remembering that 𝑣 is always the final speed at the later time 𝑡. Looking at the first second, this means 𝑣=18/ms and 𝑣=20/ms. Therefore, we have Δ𝑣=18/20/=2/msmsms. It is important to make sure 𝑣 and 𝑣 are the correct way around and not to forget the negative sign, as this indicates that the speed is decreasing. A useful tip is to check that Δ𝑣 has the expected sign for the problem: positive for increasing speed and negative for decreasing speed.

We can now calculate the acceleration in this first interval as 𝑎=Δ𝑣Δ𝑡=18/20/10=2/1=2/.msmsssmssms

So, the acceleration over the first interval is 2 m/s2.

We can calculate the acceleration over the second interval in the same way, which gives 𝑎=Δ𝑣Δ𝑡=16/18/21=2/1=2/.msmsssmssms

And in the final second, we have 𝑎=Δ𝑣Δ𝑡=14/16/32=2/1=2/.msmsssmssms

So, the car’s acceleration is 2 m/s2 throughout its motion, which means that this car has uniform acceleration of 2 m/s2.

Example 5: Understanding Acceleration

When an object is accelerating, what is happening to its speed?

  1. Its speed is decreasing.
  2. Its speed is increasing.
  3. Its speed is either increasing or decreasing.

Answer

This example asks about the meaning of the term acceleration. Recall that acceleration is the change in speed per unit time. In order to be accelerating, an object’s speed must be changing.

The problem provides three options for the change in speed. The first is that an object is accelerating when its speed is decreasing. This is true, as the speed decreasing does mean that the speed is changing. However, it is not complete: the object’s speed could also be increasing.

The second option is that an object is accelerating when its speed is increasing. Again, this is true but incomplete, as the speed could also be decreasing.

The final option is that an object is accelerating when its speed is either increasing or decreasing. This is the best answer, as it covers all of the possibilities. Option C is therefore the correct answer.

Key Points

  • Acceleration is the change in speed per unit time.
  • Acceleration can be calculated from the equation 𝑎=Δ𝑣Δ𝑡, where Δ𝑣 is the change in speed and Δ𝑡 is the change in time, or the equation 𝑎=𝑣𝑣𝑡𝑡, where 𝑣 is the initial speed, 𝑣 is the final speed, 𝑡 is the starting time, and 𝑡 is the ending time of the acceleration.
  • Acceleration can be measured in units of metres per second squared.
  • If an object’s speed changes by the same amount in equal time intervals, it has uniform acceleration.
  • An object can also have negative acceleration if its speed is decreasing.

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