Video Transcript
An athlete crosses a
30.0-meter-wide river by swimming perpendicular to the water current at a speed of
0.500 meters per second relative to the water. He reaches the opposite side at a
distance 75.0 meters downstream from his starting point. How fast is the water in the river
flowing with respect to the ground? What is the speed of the swimmer
with respect to a friend at rest on the bank of the river?
We want to solve in this exercise
for two things: first, the speed of the water in the river with respect to the
ground. We’ll call that 𝑣 sub 𝑤. Second, we want to solve for the
speed of the swimmer with respect to a stationary friend on the bank. We’ll call that speed 𝑣 sub
𝑠.
We’re told the river is
3.0
[30.0] meters wide and that the swimmer swims at 0.500 meters per
second relative to the water and also that he ends up at distance 75.0 meters
downstream from his starting point.
Let’s draw a diagram of this
situation to start our solution.
We have a river that’s 30.0 meters
wide. We can call that width 𝑤. Our swimmer sets out from the left
bank swimming at a speed we’ve called 𝑣 with respect to the water, where 𝑣 is
0.500 meters per second. If the river had no current, the
swimmer would end up on the opposite bank directly across from where he started. But there is a current in the river
moving with a speed we’ve called 𝑣 sub 𝑤. This affects the trajectory of the
swimmer so that the actual path he follows is a diagonal line. And he reaches the far shore at
distance we’ve called 𝑑 below the horizontal line of where he started. That distance is 75.0 meters. Knowing all this, we want to solve
for the current speed of the river.
If we were to zoom in on the
swimmer, we would see that there are two velocity vectors affecting the swimmer’s
motion: one we’ve called 𝑣, which is the swimmer’s self-powered motion across the
river; the other is the river current pushing the swimmer downstream. To solve for the speed of the
current, we want to remember an important fact about motion in multiple
directions. And that is that motion in
perpendicular directions is independent. In our case, this means we can
figure out how long it takes the swimmer to cross the river just by looking at the
swimmer’s own velocity 𝑣. That time to cross the river is
unaffected by 𝑣 sub 𝑤 because 𝑣 sub 𝑤 is acting perpendicular to 𝑣.
If we recall that for a constant
speed, the speed is equal to the distance travelled divided by the time it took to
travel that distance, we can write that 𝑣 — the swimmer’s speed — is equal to 𝑤 —
the width of the river — divided by 𝑡 — the time it takes for the swimmer to cross
the river. That means that the time to cross
is equal to 𝑤 divided by 𝑣 or 30.0 meters divided by 0.500 meters per second,
which is equal to 60.0 seconds. That’s how long it takes the
swimmer to reach the other side. We can reuse this expression 𝑣
equals 𝑑 over 𝑡 now to solve for 𝑣 sub 𝑤. 𝑣 sub 𝑤, the speed of the
current, is equal to the distance downstream the swimmer ended up on the opposite
side divided by the total time it took to cross the river. 75.0 meters divided by 60.0 seconds
is equal to 1.25 meters per second. That’s how fast the current in the
river is flowing.
Then, we imagine that a friend of
the swimmer is standing still on shore and watching as the river is crossed. We want to solve for the speed of
the swimmer relative to that stationary friend. If we look back at a diagram of the
velocity vectors acting on the swimmer, we see that we can combine them to form a
net velocity vector, which we’ve called 𝑣 sub 𝑠. It’s the addition of 𝑣 and 𝑣 sub
𝑤 as vectors. 𝑣 sub 𝑠 is a speed. That is, it’s the magnitude of the
addition of 𝑣 and 𝑣 sub 𝑤. So to find 𝑣 sub 𝑠, we’ll take 𝑣
sub 𝑤, square it, add it to 𝑣 squared, and then take the square root of that
sum. When we plug in the given value for
𝑣 and the value for 𝑣 sub 𝑤 we solved for and enter this expression on our
calculator, we find that 𝑣 sub 𝑠, to three significant figures, is 1.35 meters per
second. That’s the speed of the swimmer
under the influence of the river current relative to a stationary observer on
shore.
