mr.p: Good morning, today we are going to use
a toy car and a large piece of paper to do
an introductory relative motion problem.
Bo: Hi, guys.
Billy: Hey, Bo. Bobby: Hi, Bo.
♫ Flipping Physics ♫
mr.p: Bobby, could you please read
the problem just through Part A?
We'll get to Parts B and C later today.
An Introductory Relative Motion Prolem?
Bobby: Sure.
A toy car travels at 46 millimeters per second
North relative to a piece of paper that is moving
at 75 millimeters per second
West relative to the Earth.
Part A, what is the velocity
of the toy car relative to the Earth?
mr.p: Now Bobby, could you please read again?
And Bo, could you please translate?
Bobby: A toy car travels at 46 millimeters per second North
relative to a piece of paper that--
Bo: Stop please.
The velocity of the toy car with respect to the paper is
46 millimeters per second North.
mr.p: Yes, we write V sub c and then p for the velocity
of the car with respect to the paper.
Bobby, please continue.
Bobby: That is moving at 75 millimeters per second
West relative to the Earth.
Bo: Stop.
Please write V sub p and then E for the velocity
of the paper with respect to the Earth
and it's equal to 75 millimeters per second West.
mr.p: Bobby, please keep going.
Bobby: Part A, what is the velocity
of the toy car relative to the Earth?
Bo: V sub c and then E for the velocity of the car
with respect to the Earth equals question mark.
mr.p: Great, now let's take a moment to visualize the problem.
We know the car moves at 46 millimeters per second
North with respect to the paper.
In other words, if the car were not on top of the paper,
it would simply travel Northward and the paper is moving
at 75 millimeters per second Westward.
Please take a moment and point in the general direction
you think the car is going to move
when we place it on the paper.
You are all correct!
It should move in a direction
that it's somewhere between North and West.
Let's put the car on the paper and see what happens.
And you can see that we are correct.
The car moves somewhere between North and West.
Now, let's add our tip-to-tail vector diagram.
You could see the velocity of the car with respect
to the paper plus the velocity of the paper with respect
to the Earth is equal to the velocity of the car
with respect to the Earth which is, of course, an equation.
Again, the velocity of the car with respect
to the Earth is equal to the velocity of the car
with respect to the paper plus the velocity
of the paper with respect to the Earth.
And just like the last problem we did, the common subscript
either on the inside or the outside drops out and you end up
with the velocity of the car with respect to the Earth.
Bobby, could you please solve for that?
The velocity of the car with respect to the Earth.
Bobby: This looks exactly like the problem we did
when you introduced Vector Addition, is it?
mr.p: Absolutely, Bobby.
You are correct.
And I've done that on purpose actually
to show that Relative Motion is Vector Addition.
Now please solve Part A.
Bobby: We have a right triangle
so we can use the Pythagorean Theorem
or a squared plus b squared equals c squared.
So the velocity of the car with respect
to the paper squared plus the velocity of the paper
with respect to the Earth squared equals the velocity
of the car with respect to the Earth squared.
Take the square root of both sides to get the velocity
of the car with respect to the Earth is equal
to the square root of the velocity of the car
with respect to the paper squared plus the velocity
of the paper with respect to the Earth squared.
We have those numbers, so it's the square root
of 46 squared plus 75
squared, which works out to be
87.983 millimeters per second.
With two sig figs, that is 88 millimeters per second.
The velocity of the car with respect
to the Earth is 88 millimeters per second.
mr.p: The velocity of the car with respect
to the Earth is 88 millimeters per second.
Bobby: Uh...
mr.p: The velocity of the car is 88 millimeters per second.
Bobby: Oh man, I forgot direction.
Billy: That's right, velocity is a vector.
Bo: And vectors have both magnitude and direction.
mr.p: Yep, velocity is a vector, has both
magnitude and direction.
So Billy, please figure out theta.
Billy: We know tangent of theta equals opposite over adjacent
which means it will be equal to the velocity of the paper
with respect to the Earth divided by the velocity
of the car with respect to the paper.
Taking the inverse tangent of both sides gives us
theta equals the inverse tangent of the velocity
of the paper with respect to the Earth divided by
the velocity of the car with respect to the paper.
We can substitute in numbers to get the inverse tangent
of 75 divided by 46 which works out
to be 58.478 degrees.
With two sig figs, that's 58 degrees,
which means the velocity of the car
with respect to the Earth is 88 millimeters per second
at an angle of 58 degrees...
I, (sighs) I don't remember how to do
the cardinal direction thing.
mr.p: That's okay, I'll help you figure it out.
Just remember that the easiest way to do it is
to draw a compass rose on the tail of your vector.
Theta is the direction West from North.
So it's West of North.
So the velocity of the car with respect to the Earth is
approximately 88 millimeters per second
at an angle of 58 degrees West of North.
And you can see that direction
matches well to our demonstration.
Now, let's work on Part B.
Bobby, Bo, please.
Bobby: Part B, if the width of the paper necessary
for the toy car to cross is 69.2 centimeters,
how far did the toy car actually travel?
Bo: So, something equals 69.2 centimeters North,
and something else equals question mark.
I know they're both displacements,
but I don't really know what variables to use.
Maybe delta y at 69.2 centimeters North
because the width of the paper is in the y direction?
mr.p: Correct, the displacement in the y direction is
69.2 centimeters and that is North.
And we're looking for the displacement
along the hypotenuse which I have labeled delta d.
Bo, could you please find
the displacement along the hypotenuse?
Bo: We need to draw a new triangle with the displacements.
Bobby: Yeah, and it'll be a similar
triangle to the first one we drew.
Billy: And we need to convert the displacement in the y direction
to millimeters so the dimensions are the same.
So multiply the displacement in the y direction
by one meter over 100 centimeters
and then by 100, or I'm sorry, 1000 millimeters
over one meter, and we get 692 millimeters North
for the displacement in the y direction.
mr.p: Let's talk about the similar triangles for a moment.
We have a triangle with all velocity vectors
and we have a triangle with all displacement vectors.
Please, never try to put different dimensions
on the same triangle of vectors.
You need to make sure that the dimensions are the same
so that they will always play well together.
Bo, please continue.
Bo: Let's see, because they're similar triangles,
they will have the same value for cosine.
[Mr. P] Both for cosine of theta.
Bo: So the cosine of theta equals adjacent over hypotenuse
which equals the velocity of the car with respect
to the paper divided by the velocity of the car with respect
to the Earth which is then equal to the displacement
in the y direction divided by the hypotenuse displacement.
We're solving for delta d so multiply both sides
by the hypotenuse displacement which will cancel out
delta d on the right hand side.
Then we can multiply by inverse of the two velocities.
Cancel out the velocity on the left hand side
and we get that the hypotenuse displacement equals
the y displacement times the velocity of the car
with respect to the Earth, all divided by the velocity
of the car with respect to the paper.
With numbers, that's 692 times
87.983 all divided by 46 which is
1323.57 millimeters.
With two sig figs, that's 1300 millimeters.
mr.p: And who can explain to me how 1300 millimeters is
the correct answer even though it does not have a direction?
Bo: Part B asks for how far the car traveled.
And how far means the distance traveled,
not the displacement.
Bobby: And we figured out the magnitude of the displacement
which is the same as the distance traveled
because the car was moving in a straight line.
mr.p: And now Part C.
Bobby and Bo, please.
Bobby: Part C, how long did it take
the toy car to cross the paper?
Bo: Change in time equals question mark.
mr.p: Billy, it's your turn.
Billy: We know the car is moving at a constant velocity,
so let's use that equation.
Velocity equals displacement over change in time.
Substituting in our variables, we get that the velocity
of the car with respect to the Earth equals the displacement
along the hypotenuse divided by the change in time.
Solving for the change in time gives us the change in time
equals the displacement along the hypotenuse divided by
the velocity of the car with respect to the Earth.
With numbers, that is 1323.57 millimeters
divided by 87.983 millimeters per second.
The millimeters, cancel out, and we get--
Bobby: 15.043 seconds which rounds
to 15 seconds with two sig figs.
mr.p: And notice that it doesn't actually matter
which side of the triangles you choose to use
as long as you use the same side of each triangle.
We could choose the vertical sides of the triangles
and we end up with the velocity
of the car with respect to the paper
is equal to the displacement in the y direction
divided by the change in time.
When you solve that, you actually get the exact same time.
That completes the problem.
Thank you very much for learning with me today.
I enjoyed learning with you.
(chime rings)