# Video: Determining the Vector Component Form of a Velocity Resulting from an Acceleration

A spaceship is traveling at a constant velocity 𝐯 = 250.0𝐢 m/s when its rockets fire, giving it an acceleration 𝐚 = (3.0𝐢 + 4.0𝐤) m/s². What is the spaceship’s velocity 5.0 s after the rockets fire?

07:17

### Video Transcript

A spaceship is traveling at a constant velocity 𝐯 equals 250.0 meters per second in the 𝐢 direction when its rockets fire, giving it an acceleration, 𝐚, equals 3.0𝐢 plus 4.0𝐤 meters per second squared. What is the spaceship’s velocity 5.0 seconds after the rockets fire?

Now this problem is really a two-stage problem, so let’s draw a diagram showing which stages we’re talking about and how we’d go towards our final answer. So we start out with our spaceship. And our spaceship is moving ahead at a constant velocity that we’re given in the problem statement. And then we get to a certain point where the rocket engines kick on and we start to accelerate.

In that second stage of our journey, while our engines are on, we have an increasing velocity. And what we want to solve for here is the final velocity of the spaceship. We’ll call it 𝐯 sub 𝑓.

Now for the first stage of our journey when our velocity was constant, we were moving at the velocity given as 𝐯. And for the second stage, when our velocity was increasing, we have a velocity that we can call 𝐯 sub 𝑎, so the velocity when the spaceship was accelerating and this velocity is a function of time. So in the end, 𝐯 sub 𝑓, the velocity we’re trying to solve for is equal to our initial velocity 𝐯 plus our accelerating velocity or our velocity under acceleration when the time is equal to 5.0 seconds.

Since we’re already given 𝐯, our task is to solve for 𝐯 sub 𝑎 when time is equal to five seconds. So let’s turn to that challenge next. So we want to solve for 𝐯 sub 𝑓 and we’re given 𝐯, the initial velocity of the spaceship, as well as 𝐚, its acceleration when the rocket’s engines are on. Now given velocity and acceleration, a definition relating these two things may come to mind. You may recall that acceleration is defined as a change with Δ being our Greek letter representing change, a change in velocity divided by change in time. But you see here, we were given acceleration, and we want to solve for 𝐯. So how would we go about doing that?

We can start by writing out this equation, this definition, in differential form. So let’s go ahead and do that now. Okay, so here we go. Acceleration is equal to d𝐯, d𝑡: the derivative of velocity with respect to time. But what we want is not the derivative of a velocity, we want to solve for velocity itself. So to get there, let’s integrate both sides of our equation here with respect to time.

So having done that, you can see what happens here. On the right-hand side of our equation, we’re taking a derivative of velocity with respect to time as well as an integral with respect to time. So those actions cancel one another out and we’re left simply with velocity. Now that’s good news, because that’s what we want to solve for. Now look on the left-hand side of this equation. We have an integral of 𝐚, our acceleration, with respect to time. As we refer back to the given value of the acceleration we have, we can see in this equation that it does not depend on time. It’s time independent, which means we can simply multiply this acceleration, 𝐚, in our integral by the overall change in time experienced, which is 5.0 seconds.

If we do that, then we’ll be able to solve for the change in velocity that the spacecraft experiences over those five seconds. So let’s do that now. The velocity we’re solving for here is the velocity after the spacecraft has accelerated, so we’ll label it 𝐯 sub 𝑎. Now, that velocity when time equals 5.0 seconds is equal to our acceleration times that change in time, 5.0 seconds. Now we can plug in for our acceleration according to the form given to get numbers to our answer here. So let’s do that.

So here we have it. We’ve plugged in 3.0𝐢 plus 4.0𝐤 for acceleration. Now take a moment to look at the units that we’ve got here as a result of this multiplication. We have meters per second squared for our acceleration term, and we’re multiplying that by 5.0 seconds. So you see, the seconds in the numerator cancels out and our squared term cancels out in the denominator, which leaves us with units of meters per second, the same as the units of velocity. So this is very encouraging. It points to the fact that we’re on the right track with our solution. Now let’s multiply through this 5.0 seconds term to find our accelerated velocity after the spaceship has been moving for 5.0 seconds.

And that does it. So our acceleration over this five seconds give us an increase in velocity of 15𝐢 plus 20.0𝐤. So now what we wanna do now that we’ve solved for our 𝐯 sub 𝑎 term is we want to add that to our initial velocity to solve for 𝐯 sub 𝑓, the final velocity of the spaceship.

So now if we take this equation here and plug in the values we know for 𝐯 and 𝐯 sub 𝑎, we find that the final velocity of our spacecraft — Now the important thing to notice here is that we need to be careful to add within our component range, so we have two 𝐢 components and just one 𝐤 component. And remember these are like ingredients that don’t mix. They’re motion in different dimensions, so we wanna be careful to keep them separate and treat them each in their own individual category. So if you look at the 𝐢 component first, we see we’ve got 250.0 plus 15. Adding those gives us 265.0𝐢. And in the 𝐤 direction, we still have a magnitude of 20.0. So in all, the final velocity of our spacecraft is equal to 265.0𝐢 plus 20.0𝐤 meters per second.