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Question Video: Rate of Velocity Change Physics • 9th Grade

An initially stationary car starts to drive forward. After 2.5 seconds, the car has a velocity of 11.5 m/s. What is the car’s average forward acceleration?

04:21

Video Transcript

An initially stationary car starts to drive forward. After 2.5 seconds, the car has a velocity of 11.5 meters per second. What is the car’s average forward acceleration?

Okay, so in this question, we have a car that we are told is initially stationary. If the car is stationary, then this means that it’s not moving. Or in other words, we can say it has a velocity 𝑣 one of zero meters per second. Since we’re told that this is initially the case, we’ll say that this occurs at a time 𝑡 one equal to zero seconds. We are then told that the car starts to drive forward, and after 2.5 seconds, it has a velocity of 11.5 meters per second. So we can say that at the later time 𝑡 two equal to 2.5 seconds, the car now has a velocity 𝑣 two equal to 11.5 meters per second.

Now, we should keep in mind that velocity is a vector quantity. This means that it has a direction as well as a magnitude. In this case, we’re told that the car starts to drive forward, so this means that the car’s velocity is in the forward direction. Now, we could represent this by writing out explicitly in text that this velocity of 11.5 meters per second is in the forward direction. Alternatively, we can represent that direction with an arrow like this.

The question is asking us to work out the average forward acceleration of the car. We can recall that if an object changes its velocity by an amount Δ𝑣 and it does this over a time Δ𝑡, then the average acceleration 𝑎 of that object is given by Δ𝑣 divided by Δ𝑡. The change in velocity of the car is going to be equal to the car’s final velocity 𝑣 two minus its initial velocity 𝑣 one. So we have that Δ𝑣 is equal to 𝑣 two minus 𝑣 one. And subbing in our values for 𝑣 two and 𝑣 one, this gives 11.5 meters per second minus zero meters per second, which is simply 11.5 meters per second. The car changes its velocity over a time Δ𝑡, which is equal to the final time 𝑡 two minus the initial time 𝑡 one.

Subbing in our values for 𝑡 two and 𝑡 one, we get that Δ𝑡 is equal to 2.5 seconds minus zero seconds, which is 2.5 seconds. Now that we have values for both Δ𝑣 and Δ𝑡, we can substitute them into this equation to calculate the acceleration 𝑎. When we do this, we find that 𝑎 is equal to 11.5 meters per second, that’s our value for Δ𝑣, divided by 2.5 seconds, that’s our value for Δ𝑡. Evaluating this expression gives a result of 4.6 meters per second squared.

Now, just like velocity, acceleration is also a vector quantity. So in order to fully specify the acceleration of the car, we should state that this acceleration is in the forward direction. However, if we look back at the question again, we see that we are asked for the average forward acceleration. In other words, we are asked for the amount of acceleration in the forward direction, which means that we don’t actually need to include this forward direction in our answer because it’s already implicitly there.

It’s also worth making it clear that this value of 𝑎 that we’ve calculated is an average acceleration. We don’t know precisely how the car’s velocity changes as a function of time. All we know is the total change in velocity and the total time over which that change occurs. And so, this acceleration that we calculate is the average acceleration during that time. So our answer to the question is that the car’s average forward acceleration is equal to 4.6 meters per second squared.

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