### Video Transcript

A swimmer of mass 45 kilograms
uses her legs to push herself away from a swimming pool wall, applying a force
of 280 newtons. The water that the swimmer
accelerates through applies 160 newtons of force in the opposite direction to
that in which she accelerates. What is the swimmer’s
acceleration through the water?

Okay, so in this question, we’ve
got a swimmer moving through some water. So here’s our swimmer. And we’ve been told that she uses
her legs to push herself away from a swimming pool wall. So let’s say that this is the wall
that she’s pushing off from. And just for fun, we can draw in
the surface of the water.

Anyway, so we’ve been told that the
swimmer firstly has a mass of 45 kilograms. Let’s label her mass 𝑚 and say
it’s 45 kilograms. We’ve also been told that the
swimmer pushes herself off the wall, applying a force of 280 newtons. In other words, the swimmer is
exerting a 280-newton force on the wall. But what this means is that we can
recall Newton’s third law of motion.

What Newton’s third law of motion
tells us is that if object A exerts a force on object B, then object B exerts an
equal and opposite force on object A. Now in our situation here, the
swimmer is object A. It’s exerting a 280-newton force on
object B, which is the wall. Therefore, Newton’s third law of
motion tells us that the wall, object B, will exert an equal, 280 newtons, and
opposite, to the right as we’ve drawn it, force on the swimmer, 280 newtons.

Now this is important because, as
well as this 280-newton force on the swimmer to the right as we’ve drawn it, we’ve
been told that the water exerts a 160-newton force on the swimmer in the opposite
direction to that in which she accelerates. Well, the swimmer is accelerating
to the right because she’s trying to push away from the wall. And of course, once again, we’ve
just arbitrarily drawn the wall to the left of the swimmer. We could’ve equally drawn the
swimmer going in the opposite direction and the wall to the right. But the point is that the swimmer
is trying to accelerate away from the wall.

And so we can say that her
acceleration is to the right as we’ve drawn it. And we’ll call this acceleration
𝑎. But then the water we’ve been told
exerts a 160-newton force on the swimmer in the opposite direction to which she’s
accelerating. This means that there are now two
forces acting on the swimmer: the 280-newton force to the right because of the wall
and 160-newton force to the left because of the water.

We can simplify all of this by
drawing the swimmer just as a blob and considering that there’s a 280-newton force
to the right and a 160-newton force to the left. Then we can account for these two
forces and find the resultant or overall force on the swimmer. This overall force is also called
the net force. And if we consider arbitrarily that
the direction to the right is positive and the direction to the left is negative,
then we can add these two forces together.

And the net force, which we’ll call
𝐹 subscript net, is equal to the positive 280 newtons, cause that force is to the
right, minus the 160 newtons, which is to the left. And that net force, therefore, ends
up being 120 newtons. Therefore, it’s a 120-newton force
to the right. And this kind of makes sense. If there’s a 280-newton force to
the right and a 160-newton force to the left, then 160 of those 280 newtons to the
right are being cancelled by this force to the left. And whatever remains — that’s the
120 newtons — is going to be the resultant force to the right.

If, however, we had arbitrarily
chosen the left to be positive and the right to be negative, then we’d have found
the net Force, 𝐹 subscript net, to be the positive 160 newtons minus the 280
newtons. And that would’ve given us a net
force of negative 120 newtons. So what gives?

Well, the reason for this is
because we chose this way, to the left, to be positive. And hence, what we found is that
the net force is negative 120 newtons to the left. Or in other words, it’s 120 newtons
to the right, exactly as what we found here. So even though we do need to be
careful with signs, regardless of which convention we chose, we’d find the net force
to be 120 newtons to the right.

So now that we found the net force
on our swimmer, we can recall Newton’s second law of motion. This law tells us that the net
force on an object is equal to the mass of that object multiplied by the
acceleration it experiences. And since we’ve already found the
net force and we’ve been given the mass of the swimmer, we can find out the
acceleration experienced by the swimmer.

We can do this by rearranging the
equation by dividing both sides of the equation by the mass, which results in the
mass cancelling out on the right. This way, what we’re left with is
that the net force on the swimmer divided by the mass of the swimmer is equal to the
acceleration experienced by the swimmer. So we can say that the
acceleration, 𝑎, is equal to the 120-newton net force to the right divided by the
mass of the swimmer, which is 45 kilograms.

We can also see that we’re working
in base units because the base unit of force is the newton and the base unit of mass
is the kilogram. And so our answer for the
acceleration is going to be in its own base unit, which is metres per second
squared. And so when we evaluate this
fraction on the right-hand side, we find that the acceleration is equal to 2.6
recurring. And the unit is metres per second
squared.

Now because the smallest number of
significant figures that we’ve been given in the question is two significant
figures, we can give our final answer to two significant figures as well. Now 2.6 recurring is basically two
point and then six occurring many times. And the reason we write it like
this is so that we can work out that this is the first significant figure and this
is the second.

Well, the second significant figure
is either going to stay the same or it’s going to round up, depending on what the
next significant figure is. The next one is a six. Since six is larger than five, this
second significant figure is going to round up. And so our value becomes 2.7 metres
per second squared. So at this point, we have our final
answer. The swimmer’s acceleration is 2.7
metres per second squared.