### Video Transcript

A van that has a mass of 750 kilograms increases its momentum by 18750 kilograms meters per second when it accelerates at a constant 2.5 meters per second squared. For how long does the van accelerate?

Alright so in this question, we know that we’ve got a van. So here’s our van. And we’ve been told that it has a mass, which we’ll call 𝑚, of 750 kilograms. Now, we also know that the van is accelerating at a constant rate of 2.5 meters per second squared. So let’s say that the acceleration of the van, which we’ll call 𝑎, is 2.5 meters per second squared.

As a result of this acceleration, we’ve been told that the van increases its momentum by 18750 kilograms meters per second. So from here, the position the van is at currently, to the position that the van ends up in, there is a change in the momentum of the van. We’ll call this change in momentum Δ𝑝 because Δ represents change and 𝑝 represents momentum. And this change in momentum is 18750 kilograms meters per second.

Now, what we’ve been asked to do is to find out for how long the van accelerates. In other words, we’ve been asked to find the time interval between this position and this position. So let’s say that the van starts accelerating at a time 𝑡 one and finishes accelerating at a time 𝑡 two. We’ve been asked to find 𝑡 two minus 𝑡 one or, in other words, the time interval over which the acceleration happens. Or a simpler way to write this is simply Δ𝑡, the change in the time over which the acceleration happens. At which one we can forget about 𝑡 two and 𝑡 one because we don’t need to know at exactly what time the van starts accelerating and finishes accelerating.

All we need to find is the time interval, the amount of time over which the van accelerates. So how are we going to go about doing this? Well, we’re going to start by recalling Newton’s second law of motion. Newton’s second law of motion tells us that the force on an object is equal to the mass of the object multiplied by the acceleration experienced by the object. Now in this question, we’ve been given the mass of the van and the acceleration of the van. And so we would be able to calculate the force on the van.

However, is this actually any use to us? Well actually, the reason that we’ve recalled Newton’s second law of motion is so we can take a closer look at the right-hand side of this equation. The reason is that we can recall the definition of acceleration. Acceleration 𝑎 is defined as the rate of change of velocity, in other words, the change in velocity of an object divided by the time interval over which this change in velocity occurs. But then if that is the definition of acceleration, we can say that the mass of an object multiplied by its acceleration is equal to the mass of the object multiplied by Δ𝑉 divided by Δ𝑡 because that whole thing is acceleration.

But then at this point, we can recall that momentum, 𝑝, is defined as the mass of an object multiplied by the velocity of an object. Now, why is this useful? Have we just introduced another equation for no reason? Well no, if the momentum is equal to the mass of an object multiplied by its velocity, then the change in momentum which we’ll call Δ𝑝 — notice, by the way, that Δ𝑝 is one of the quantities we’ve been given in the equation. But then we can say that Δ𝑝 is equal to the change in mass times velocity.

Now at this point, we need to realize that for any object that has a constant mass such as for example our van, the mass of the van is not changing over its acceleration period. For an object like this, the change in mass times velocity is the same thing as mass multiplied by the change in velocity because the mass is constant. So we can pull it out of the parenthesis. And this Δ representing changing will only affect the velocity. This is because the only thing that can change is the velocity. And hence, we can say that the change in momentum Δ𝑝 is equal to the mass of an object multiplied by the change in velocity of an object if and only if the mass is constant which, in this case, it is.

So why are we doing all of this algebraic gymnastics? Well, the reason for this is that we can see 𝑚 multiplied by Δ𝑉 over here in the equation that we wrote earlier. We’ve got 𝑚 multiplied by Δ𝑉 in the numerator. And so we can replace that with Δ𝑝, the change into momentum. And hence, what we’re left with is that the mass of an object multiplied by its acceleration is equal to the change in momentum of the object divided by the time interval over which that change occurs. But of course, it’s only true if the mass is constant which, in this case, it is.

And at this point, we’ve arrived at an equation where we know the value of 𝑚; that’s the mass of the van. We know the value of 𝑎; that’s the acceleration of the van. We also know the value of Δ𝑝; that’s the change in momentum of the van. And we don’t know the value of Δ𝑡 which is exactly what we’re trying to find. So now, all we need to do is to rearrange our equation. We can start by multiplying both sides of the equation by Δ𝑡 so that the Δ𝑡 on the right-hand side cancels.

This way, what we have is 𝑚𝑎 multiplied by Δ𝑡 is equal to Δ𝑝. Then we divide both sides of the equation by 𝑚𝑎. This way, 𝑚𝑎 on the left-hand side cancels. And what we’re left with is that Δ𝑡 is equal to Δ𝑝 divided by 𝑚𝑎. Now, we can substitute in the values that we’ve been given. What we get is that Δ𝑡 is equal to 18750 kilograms meters per second, that’s the change in momentum, divided by 750 kilograms, that’s the mass of the van, and 2.5 meters per second squared, that’s its acceleration. Quickly starting the units, we see that the unit of kilograms in the numerator cancels with this kilograms in the denominator.

A further unit of meters in this numerator cancels with meters in the denominator. And then one power of per second cancels with one power of per second in the denominator. So overall, what we’re left with is one divided by seconds in the denominator which is equivalent to having seconds in the numerator. In other words, our final answer when we evaluate this fraction is going to be in seconds. And this is a good thing because Δ𝑡 is a time interval. So we need it to be in seconds. So evaluating the fraction on the right-hand side of the equation, we find that Δ𝑡, the time interval, is equal to 10 seconds. And hence, we have the answer to our question. The van accelerates for 10 seconds.