Video: Applying Galilean Transformations of Velocity to Perpendicular Velocities

A man is running on a straight road that is perpendicular to a train track; he runs away from the track at a speed of 12 m/s. The train moves at 30 m/s with respect to the track. What is the speed of the man with respect to a passenger sitting at rest in the train?

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Video Transcript

A man is running on a straight road that is perpendicular to a train track. He runs away from the track at a speed of 12 metres per second. The train moves at 30 metres per second with respect to the track. What is the speed of the man with respect to a passenger sitting at rest in the train?

Before we go and draw a diagram of this situation, let’s highlight some important information in this problem statement. So we see that the man is running away from the track at a speed of 12 metres per second and that the train is moving at 30 metres per second relative to the track. We can give names to these speeds. Well, we’ll assign each one a variable. So let’s call the speed of the man 𝑉 sub π‘š. And we can call the speed of the train relative to the track 𝑉 sub 𝑑.

Now, let’s go ahead with our diagram. So here in this drawing, we have a dot representing the man and his motion along the track, and again we know that that is 12 metres per second. And as well we have a picture of the train or a sketch of the train moving along the track, and we’ve been given that speed as 30 metres per second. In the question that we want to figure out is β€œwhat is the velocity of the man relative to a passenger who is sitting still on the train?” See the passengers seated there on the train. So in other words, what we wanna know is the relative velocity, which we can call 𝑉 sub π‘Ÿ of the man running along the track, compared to a passenger sitting still on the train as it moves along its track.

Now because we’ve drawn the man and the train’s velocity vectors as arrows, you might be thinking that we could put them on a coordinate plane. And that’s exactly right. If we put them on a plane, then that will help us figure out what is 𝑉 sub π‘Ÿ, their relative velocity. Now when we sketch these vectors out on a 𝑉 sub π‘₯ and 𝑉 sub 𝑦 coordinate plane, if we put the vectors tip to tail, then that allows us to set it up graphically. So we can solve for the resultant velocity, 𝑉 sub π‘Ÿ.

The green vector that I’ve drawn in there, that vector represents the relative speed 𝑉 sub π‘Ÿ of the man running along the road relative to a passenger seated on the train. So the question is: what is 𝑉 sub π‘Ÿ? Well as we look at this triangle, we see that it is a right triangle meaning the Pythagorean theorem will let us solve for the third leg of the triangle that is unknown, given the fact that we already know two of the legs. So let’s write this theorem down before we start.

That theorem tells us that in a right triangle with sides π‘Ž, 𝑏, and 𝑐, where 𝑐 is the longest side, the lengths relate to one another. This way, π‘Ž squared plus 𝑏 squared equals 𝑐 squared. When it comes to our scenario, we can write that this way: 𝑉 sub 𝑑 squared plus 𝑉 sub π‘š squared is equal to 𝑉 sub π‘Ÿ squared. So you can probably see that now we wanna take the square root of both sides of our equation to isolate 𝑉 sub π‘Ÿ.

And having done that, the square root and the square cancel out and we’re left with an expression that says the square root of 𝑉 sub 𝑑 squared plus 𝑉 sub π‘š squared is equal to 𝑉 sub π‘Ÿ, the resulting velocity. Now, it’s time to plug in our values for 𝑉 sub π‘š and 𝑉 sub 𝑑. So 𝑉 sub π‘Ÿ is equal to the square root of 12 metres per second squared plus 30 metres per second squared.

When we square those terms out, we find 12 squared is 144 and 30 squared is 900. Adding those terms together, we see 𝑉 sub π‘Ÿ is equal to the square root of 144 metres squared per second squared. That number is equal to 32 metres per second. We keep two significant figures in our answer because we were given initial values that themselves had two significant figures. So the velocity of the man running along the road relative to a passenger sitting still on the train is 32 metres per second.

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