### Video Transcript

A rocket is flying upward at a constant speed when it ejects its empty first stage boosters and activates its second stage boosters. When this happens, the rocket has a net upward acceleration of 15 metres per second squared in its direction of travel. The second stage boosters burn for seven seconds. At the end of the burn, the rocket’s upward velocity is 250 metres per second. What velocity did the rocket have before firing the second stage boosters?

Okay, so, this is a long and complicated question with lots of information. Let’s try and break it down into smaller digestible chunks. Let’s first start with thinking about the rocket which is flying upward at a constant speed. This is what the rocket is doing initially before anything that we’re about to consider even happens.

So, here’s our little rocket. And it’s moving upwards, we’ve been told, at a constant speed, which we can call 𝑢. Now at some point, the rocket ejects, or gets rid of, its empty first stage boosters. So, let’s say that these blue bits here are the first stage boosters. And they’ve now run out of fuel. So, here’s the rocket ejecting the two first stage boosters. Now, before we do anything else, let’s write down on the side that we said that the initial velocity of the rocket was 𝑢. We can write this down so that we don’t forget it for later.

But anyway, so, we’ve been told in the question that as soon as the rocket gets rid of its first stage boosters, it also activates its second stage boosters. So, lets draw some flames coming out the back of the rocket because we can say that that’s where the second stage boosters are. Now we’ve been told that the combined effects of getting rid of the first stage boosters and activating the second stage boosters results in the rocket having a net upward acceleration of 15 metres per second squared in its direction of travel.

Well, the direction of the rocket’s travel is indeed in the upward direction. And we can draw this double arrow in the upward direction to say that the acceleration, which we’ll call 𝑎, is 15 metres per second squared. Now as well as this, we’ve been told that the second stage boosters burn for seven seconds. In other words, here’s the rocket seven seconds later. It’s moved further up, and the second stage boosters have just stopped burning.

Hence, we can say that the time interval between when the rocket had just started burning its boosters and when it finishes burning its boosters is seven seconds. Or another way to think about this, is that if we say that the rocket started burning its boosters at a time 𝑡 one and finished burning at a time 𝑡 two. Then we’ve just found out that the time interval, which we’ll call Δ𝑡, Δ representing changing and 𝑡 representing time, is equal to 𝑡 two, the final time, minus 𝑡 one, the initial time. Which is basically the time interval over which the rocket’s second stage boosters were burning. And that happens to be seven seconds.

Now another thing we’ve been told in the question is that at the end of the burn, the rocket’s upward velocity is 250 metres per second. In other words, at this point, the rocket is moving upwards at 250 metres per second. Let’s say that this is the rocket’s final velocity, and we’ll call this 𝑣. What we’ve been asked to do in the question is to find the velocity that the rocket was flying at before firing the second stage boosters. In other words, what was the velocity of the rocket just before it got rid of the first stage boosters and activated the second stage boosters?

Then we said that just before it did that, it was moving at a constant velocity 𝑢. That was its initial velocity. And now we’ve asked to find the value of 𝑢. So, how do we go about doing that? Well, we know the acceleration of the rocket between when it starts running its boosters and stops burning its boosters, as well as the time interval between those two points, and the final velocity of the rocket. Hence, we can recall the definition of acceleration.

We can recall that acceleration is defined as the change in velocity of an object, that’s Δ𝑣, Δ representing change and 𝑣 representing velocity, divided by the time interval over which this velocity change occurs. In other words, the difference in time between when the rocket starts accelerating and finishes accelerating.

Now we already know the value of 𝑎 and we know the value of Δ𝑡, the change in time, or in other words, the time interval. However, what we don’t know is the change in velocity of the object. But we do know the final velocity of the object. And so, just like we saw over here that the change in time, or the time interval, is equal to the final time minus the initial time, we can recall that the change in velocity of an object is equal to the final velocity of the object, which we’ve called 𝑣, minus the initial velocity, which we’ve called 𝑢. And so, we can substitute this into acceleration equation.

We can say that the acceleration of the rocket is equal to the final velocity minus the initial velocity divided by the time interval, which we already know. And now we’ve got an equation where we know this, we know this, and we know this. The only thing we’re trying to find is 𝑢. So, we can rearrange to find 𝑢. We can start by first multiplying both sides of the equation by Δ𝑡 so that it cancels on the right-hand side. Which means that what we’re left with is Δ𝑡 multiplied by 𝑎 on the left and 𝑣 minus 𝑢 on the right.

Then we can multiply both sides of the equation by negative one. The reason we do this is because on the right-hand side we have 𝑣 minus 𝑢, but ideally, we want 𝑢 to be positive because that’s what we’re trying to solve for. So, on the left-hand side, we have Δ𝑡 multiplied by 𝑎. And we multiply this by negative one. And so, it becomes negative Δ𝑡 multiplied by 𝑎. Or an easier way to write this is 𝑎 times Δ𝑡, or 𝑎Δ𝑡.

And so, on the left, we have 𝑎Δ𝑡. And on the right, we have negative one multiplied by 𝑣, which is negative 𝑣. And we add to that negative 𝑢 multiplied by negative one, which becomes positive 𝑢. And so, on the right, we have minus 𝑣 plus 𝑢, or 𝑢 minus 𝑣, the simpler way to write it. And so, clearing things up a bit, we have negative 𝑎Δ𝑡 is equal to 𝑢 minus 𝑣. At this point, we just add 𝑣 to both sides of the equation, which means we no longer have 𝑣 on the right-hand side. And this leaves us with 𝑣 minus 𝑎Δ𝑡 is equal to 𝑢.

So, now all we need to do is to sub in the values of 𝑣, 𝑎 and Δ𝑡, which looks like this. We have the final velocity, 250 metres per second, minus the acceleration, 15 metres per second squared, times the time interval, seven seconds. And that is going to give us the initial velocity.

Now looking at the units quickly, we have the units of metres per second over here. And in this factor, we have metres per second squared multiplied by seconds, which means that one factor of seconds is going to cancel from the numerator and the denominator. And thus, we’re left with metres per second overall. Then, when we subtract this from this, the overall unit is going to be metres per second once again. And that’s a good thing because we’re trying to find a velocity, which has units of metres per second.

So, this term on the right, 15 times seven, ends up being 105 metres per second. At which point, we have 250 metres per second minus 105 metres per second. And evaluating this, we find that 𝑢, the initial velocity, is 145 metres per second. So, our final answer is that the velocity of the rocket before firing the second stage boosters was one 145 metres per second.