### Video Transcript

A car has an initial momentum of 1500 kilogram meters per second when the driver starts to speed it up, changing its momentum to 1750 kilogram meters per second in a time of 0.75 seconds. Determine the average force that acts on the car’s wheels while it accelerates. Answer to the nearest newton.

Okay, so in this question, we have a car that we’re told starts out with an initial momentum of 1500 kilogram meters per second. Let’s suppose that this here is the car. Since momentum is a vector quantity, it has to have a direction as well as a magnitude, and the momentum of the car is going to be in the direction that it’s moving in, which in our diagram is in this direction. Let’s label the initial momentum of the car as 𝑝 subscript i so that we have 𝑝 subscript i is equal to 1500 kilogram meters per second.

We are then told that the driver starts to speed up the car and that this changes its momentum to 1750 kilogram meters per second. So let’s suppose that this is the same car after its momentum has been changed. Now its momentum is still going to be in the same direction, and we can indicate that with an arrow like this. We’ll label the new momentum, or final momentum, of the car as 𝑝 subscript f so that we have 𝑝 subscript f is equal to 1750 kilogram meters per second. We’re told that this momentum change happens over a time of 0.75 seconds. So the time interval between the car being at this position and the car being at this position is 0.75 seconds. We’ve labeled this time interval as Δ𝑡.

We are asked to work out the average force that acts on the car’s wheels while it accelerates. In order to do this, we can recall that if an object changes its momentum by an amount Δ𝑝 over a time interval of Δ𝑡, then this means that the force 𝐹 which acts on the object while its momentum changes is equal to Δ𝑝 divided by Δ𝑡. In this case, we know that during a time interval of Δ𝑡 equal to 0.75 seconds, the car’s momentum changes from an initial value that we’ve called 𝑝 subscript i to a final value that we’ve called 𝑝 subscript f. So the change in the car’s momentum Δ𝑝 is going to be equal to the final momentum 𝑝 subscript f minus the initial momentum 𝑝 subscript i.

Subbing in our values for 𝑝 subscript f and 𝑝 subscript i, we have that Δ𝑝 is equal to 1750 kilogram meters per second minus 1500 kilogram meters per second, which is 250 kilogram meters per second. So we now know the value of Δ𝑝, the amount that the car’s momentum changes by. And we also know the value of Δ𝑡, the time interval over which this momentum change happens. So now we just need to sub those two values into this equation to calculate the force 𝐹. When we do this, we find that 𝐹 is equal to 250 kilogram meters per second, that’s our value for Δ𝑝, divided by 0.75 seconds, that’s our value for Δ𝑡.

At this point, it’s worth noticing that the momentum is in units of kilogram meters per second, which is the SI base unit for momentum. And the time is in units of seconds, the SI base unit for time. Since both these two quantities are in their SI base units, this means that the force we calculate will be in its own SI base unit. The SI base unit of force is the newton, which we represent as a capital N. When we evaluate this expression, we get a result of 333.3 recurring newtons. This value is the average force that acts on the car’s wheels, which is what we were asked to find. Since the value that we’ve calculated is positive, this means that the force acts in the same direction as the car’s momentum. So in our diagram, that’s this direction here.

Looking back at the question, we see that we are asked to give our answer to the nearest newton. Rounding this result to the nearest newton, we get our answer to the question that the average force that acts on the car’s wheels while it accelerates is equal to 333 newtons.