### 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.