A car is being driven with a constant acceleration of 25 meters per second squared along a straight horizontal road. How much will the car’s velocity increase in 4.00 seconds?
The question is asking us to find the velocity increase that comes from an acceleration of 25 meters per second squared applied over a period of four seconds. To do this, we’ll need a relationship between a general velocity change which we’ll call Δ𝑣, in this case an increase, and acceleration which we’ll call 𝑎 and the time interval in which that acceleration is applied, which we’ll call Δ𝑡. A good place to start looking for this relation would be the definition of acceleration. Acceleration is the rate of change of velocity, which is another way of saying the change of velocity with respect to time.
Since velocity is a vector quantity, it has two parts, speed and direction, and acceleration is a change in one or both of these parts. Let’s illustrate this with two examples. In the first example, we’ll have a car going around a bend in the road. The car will start at point one and travel to point two. Our second example will be a car travelling along a straight road. And again, we’ll have the car start at point and travel to point two. Let’s let both cars have an initial velocity that’s the same size and same direction. Now, let’s have the cars travel to point two. On the curved road, we’ll say that the size of the velocity hasn’t changed. But the direction clearly has.
On the other hand, on the straight road, the direction hasn’t changed. But for the sake of the example, we’ll say that the size has. The important thing to note is that both of these cars have accelerated. The first car’s velocity has changed direction, even though it hasn’t changed size. And the second car’s velocity has changed size, even though it hasn’t changed direction. So both velocities have changed with time. In other words, both cars have accelerated. Remember, though, that the question talked about a straight horizontal road, which is like the second example we discussed where direction is constant. So we only have to worry about the velocity changing size.
So let’s look at how the size of velocity changes under constant acceleration. Let’s use a car again. Here’s a car with an initial velocity of 20 meters per second. Let’s let the car also have a constant acceleration of 10 meters per second squared. Meters per second squared is meters per second, per second. So every second, velocity will increase by 10 meters per second. With this acceleration, the size of the car’s velocity after one second is 30 meters per second or 10 more than it was initially. After another second elapses, the size of the car’s velocity increases by another 10 meters per second to 40 meters per second.
From this example, we can already start to see how to solve this problem since the question gives us a constant acceleration and a time interval in seconds. Before we work with the question though, let’s put this example on some more rigorous footing. Symbolically, we would write change in velocity as Δ𝑣 and time elapsed as Δ𝑡. Acceleration then being the change in velocity with the given time interval can be written as 𝑎 equals Δ𝑣 divided by Δ𝑡. Let’s check this against our example. The final size is 40 meters per second, the initial size is 20 meters per second, and the direction is the same. So the net change in velocity is 40 meters per second minus 20 meters per second is equal 20 meters per second, which is Δ𝑣.
And this final velocity was reached after two seconds. So Δ𝑡 is two seconds. Now, taking Δ𝑣 divided by Δ𝑡, we find 20 meters per second divided by two seconds is equal to 10 meters per second squared, which identically matches the value that we gave acceleration at the beginning of the example, exactly as it should. Now that we’ve seen that 𝑎 equals Δ𝑣 divided by Δ𝑡 works for a constant acceleration and when velocity doesn’t change direction, let’s apply to our problem where acceleration is constant and velocity doesn’t change direction.
We have Δ𝑣 over Δ𝑡 is equal to 𝑎. But we wanted a relationship that had Δ𝑣 on its own because that’s what we’re trying to solve for. So let’s multiply both sides by Δ𝑡. On the left-hand side, Δ𝑡 divided by Δ𝑡 is just one, and that leaves Δ𝑣 on its own on the left equal to 𝑎 times Δ𝑡 on the right. And this is the relation we’re looking for, Δ𝑣 in terms of 𝑎 and Δ𝑡. So let’s plug in and solve. 25 meters per second squared times 4.00 seconds is equal to 100 meters per second. And that’s our final answer. The car’s velocity increases by 100 meters per second.