In this video, our topic is
relative velocity. Relative velocity helps us describe
the movement of one object in motion with respect to another moving object. For example, it could help us
understand the motion of one of these race cars with respect to one of the
To get started, imagine that we
have two particles, one here and one here, and that each one of these particles is
in motion. We could imagine the particle on
the right is moving to the right at five meters a second, while the one on the left
is moving to the left at 10 meters per second. What we’ve done here is defined the
speeds of these two particles with respect to what we could call their
environment. We imagine some fixed frame, the
ground, for example, and we measured the speeds of each particle with respect to
Since we’re talking about
velocities in this lesson, which we know are vectors with both magnitude and
direction, let’s define a positive direction of motion for these two particles. We’ll say that any particle moving
to the right is moving in the positive direction. Therefore, the velocity of this
particle is indeed positive five meters per second, while the particle on the left
actually has a negative velocity. Let’s say further that these
particles are big enough that we can stand on them. From the perspective of the
particle on the left, we wonder, what’s the velocity of the particle on the
right? In other words, what’s the velocity
of this particle relative to this one where we stand?
In raising this question, we’re
essentially changing our frame of reference. Before, we defined the velocities
of these two particles relative to the frame of reference of the ground. But now we’re saying that we, as we
stand on the left particle, are that frame of reference. How then do we perceive the
velocity of the particle on the right?
Well, we know that it’s no longer
five meters per second. That was this particle’s velocity
relative to the ground, but not to us as we move away from it. The velocity we perceive for the
particle on the right, we’ll call it 𝑉 sub R, is equal to that particle’s velocity
relative to the ground minus our particle’s velocity relative to ground. Five meters per second minus
negative 10 meters per second Works out to positive 15 meters per second. So if we’re standing still on the
particle on the left, then the one on the right looks like it’s moving away from us
with a velocity of positive 15 meters per second.
But now look at this. Say that we switch particles so now
we’re standing on the one on the right. We might expect that if from this
perspective we were to calculate the velocity of the particle on the left, we would
get this answer of positive 15 meters per second. But let’s see. 𝑉 sub L is equal to the velocity
of the particle on the left relative to the ground, that’s negative 10 meters per
second, minus the velocity of the particle on which we stand, positive five meters a
second. For our answer then, we end up with
negative 15 meters per second.
What we’re seeing is that the
relative velocity between two particles depends on our vantage point. And it also depends on the vector
nature of velocity, that it has a direction.
Let’s now consider a different
scenario where here we have a boat moving across a lake. We’ll say it has a speed of 10
miles per hour. And as the boat moves, we walk
across its deck in the same direction as its motion at a speed of three miles per
hour. Just like before with our moving
particles, in this case, there are different relative velocity calculations we could
make. For example, we could solve for our
velocity relative to the boat.
If we assume, like before, that
motion to the right is positive, then that’s actually pretty easy because we’re
already told our speed relative to the boat. But say that instead we want to
calculate our velocity relative to the shore. To solve for this, we’ll need to
take the velocity of the boat relative to the shore as well as our velocity relative
to the boat into account. What we would do is add those two
velocities, positive 10 and positive three miles per hour, together so that our
velocity relative to the shoreline we’re approaching is 13 miles per hour.
So we see that, in this example,
we’ve added together our two velocities, whereas recall that in this case we were
subtracting the velocities of the two particles from one another. This helps us see that when we
calculate relative velocities, every situation must be evaluated on its own
merits. And we need to be careful to
observe any sign conventions we establish for a given scenario.
To this point, we’ve solved for
relative velocities, such as 𝑉 sub s in this example, algebraically. But it’s also possible to do this
using vectors. For our boat scenario, we could
define an axis. And say that relative to the zero
of this axis, we have a vector representing our velocity relative to the boat. And then to that we add a second
vector representing the boat’s velocity relative to shore.
Combining these vectors graphically
gives us the resultant vector. And if we think carefully, we can
take a similar approach over here with our two particles in relative motion. Say that we’re back to standing on
the particle on the left and wanting to calculate the relative velocity of that on
the right. This means we take the positive
five meter per second vector for the particle on the right and we add to it the
magnitude of the vector of our frame of reference, the particle on the left. So we add a vector of magnitude 10
meters per second, giving us the resultant of 15 meters per second.
Now let’s switch it up and say
we’re standing on the particle on the right. To solve for the relative velocity
of the particle on the left graphically, we take the velocity vector of the particle
on the left, negative 10 meters per second, and we contribute to it in that
direction our speed of five meters per second. When we add that in the negative
direction, it points in that same direction, giving us the result we found earlier,
negative 15 meters per second.
Note that, so far, we haven’t
written down any rules about how to calculate relative velocity. This is because when it comes to
relative velocity, each situation needs to be evaluated individually. An intuitive understanding of what
we’re trying to calculate relative to what frame of reference is very helpful. That kind of intuition is developed
through practice. So let’s look now at a few example
A car is moving on a straight road
at 84 kilometers per hour. And in the opposite direction, a
motorbike is moving at 45 kilometers per hour. Suppose that the direction of the
car is positive. Find the velocity of the motorbike
relative to the car.
Okay, let’s say that this is our
straight road and this is our car moving along, we’ll say, to the right at 84
kilometers an hour, which means the motorbike is moving to the left at a speed of 45
kilometers an hour. We’re told that the direction of
the car to the right is positive. And we want to solve for the
velocity of the motorbike relative to the car.
When we say relative to the car, we
mean that we’ll essentially be treating the car as though it’s stationary. And we want to solve for the
velocity of the motorbike relative to that stationary observation point. That is, if the car had a velocity
of zero, what would the velocity of the motorbike be?
Note that we’re given the speeds of
both of these vehicles relative to the stationary road. Compared to the road, the car has a
velocity of positive 84 kilometers per hour. And because it moves in a direction
we’ve defined as negative, the motorbike has a velocity relative to the road of
negative 45 kilometers an hour.
Now, like we mentioned, solving for
the velocity of the motorbike relative to the car, we can call this 𝑉 sub mc,
involves imagining that our car is stationary or has a velocity of zero and then, in
that frame, solving for the resulting velocity of the motorbike. Given the car’s velocity of
positive 84 kilometers per hour, we can see that to make this zero, we would need to
subtract that number from it and that if we were to do this, the velocity of the car
would be zero. This tells us we need to apply the
same operation to the velocity of the motorbike relative to the road to move it into
the frame of reference where the car is stationary, where it’s our reference
point. 𝑉 sub mc then is negative 45
kilometers per hour minus 84 kilometers per hour, which equals negative 129
kilometers an hour. This is the velocity of the
motorbike relative to the car.
Let’s look now at another
A ship was sailing with a uniform
velocity directly toward a port that is 144 kilometers away. A patrol aircraft passed over the
ship, traveling in the opposite direction, at 366 kilometers per hour. When the aircraft measured the
ship’s speed, it appeared to be traveling at 402 kilometers per hour. Determine the time required for the
ship to reach the port.
Okay, so let’s say that we’ve got
this ship moving across the salty sea and headed for port. That port is a distance, we’re
told, of 144 kilometers away. And while all this is going on, a
patrol aircraft flying in the direction opposite the way the ship is moving passes
over the ship and measures the ship’s speed relative to the aircraft to be 402
kilometers per hour. Based on this, we want to solve for
the time required for the ship to reach port.
As we start on our solution, let’s
recall that when an object moves at a constant speed, we can calculate that speed 𝑠
by dividing the distance the object travels by the time it takes to travel that
distance. And note that we can rearrange this
equation so that it reads 𝑡 is equal to 𝑑 divided by 𝑠.
In our scenario, it’s a time that
we want to solve for. And we’re given a distance. That’s 144 kilometers between the
ship and port. But we don’t yet know the speed of
our ship. We’ll call that speed 𝑆 sub s. And while we don’t know it, here’s
what we do know. Our patrol plane flying in the
opposite direction at 366 kilometers per hour perceives the ship to be moving at 402
kilometers per hour. In other words, if we take the
speed of the ship 𝑆 sub s and we add it to the real speed of the plane, then we’ll
get the perceived speed of the ship relative to the plane.
And note that we add these two
values rather than, say, subtract them because the ship and the plane are
approaching one another. This equation tells us that if we
subtract 366 kilometers per hour from both sides, then we’ll have that 𝑆 sub s
equals 402 kilometers per hour minus 366 kilometers an hour, which means the speed
of the ship relative to the water is 36 kilometers per hour. And this is the speed that we’ll
want to use in our equation to solve for the time required for our ship to reach
Now that we know that 𝑆 sub s is
36 kilometers an hour, we can write that the time it takes for our ship to reach
port is equal the distance it has to travel divided by 𝑆 sub s. That’s 144 kilometers divided by 36
kilometers per hour. And note that the units of
kilometers cancel out, while the units of hours will move up to the numerator. 36 goes into 144 exactly four
times. So our answer is that the ship
takes four hours to reach port.
Let’s now look at an example
involving motion in two dimensions.
A certain river is half a mile
wide, with a current flowing at two miles per hour from east to west. A man swims directly toward the
opposite shore from the south bank of the river at a speed of three miles per
hour. Find the distance that he travels
when swimming from shore to shore. How far down the river does he find
himself when he has swam across?
Okay, if we say that these are our
four compass directions, then we can say that if this is our river, with current
flowing east to west, then our man starts on the south bank and begins to swim
across to the north bank. We’re told that the man swims in
this direction, directly toward the opposite shore. But thanks to the current in the
river, we know that he won’t actually follow a straight line path like this. Rather, the current will push him
downstream more and more over time. So he follows a path looking like
this. We want to know how far the man
travels while swimming, in other words, this distance. And secondly, we want to know how
far down the river does he reach the opposite shore. That’s indicated by this length
Let’s call the total distance that
the swimming man travels 𝑑 sub t and the distance down shore he arrives 𝑑 sub
s. And notice that if we consider the
third side of what is a right triangle, this distance is given to us in the problem
statement. It’s one-half mile. We have here then this right
triangle with distances on each one of its sides. And currently we know one of those
distances and want to solve for the other two. Along with this distance
information though, we’re also given information about speeds: first that the
current is flowing two miles an hour east to west and that the man is swimming
directly across the river at three miles an hour.
We can write these two velocity
vectors in like this. And note that these also meet at a
right angle. This suggests that even though our
man is swimming directly across the river at three miles an hour, because he’s also
being pushed downstream by the current, his actual speed as he travels, we’ll call
it simply 𝑠, is greater than either the speed of the current or his three mile an
hour swimming speed.
Indeed, because this is a right
triangle, we can use the Pythagorean theorem to solve for 𝑠. Before we do that though, let’s
clear a bit of space on screen. We can recall that the Pythagorean
theorem tells us that, for a right triangle with sides 𝐴, 𝐵, and 𝐶, where 𝐶 is
the hypotenuse, 𝐶 squared equals 𝐴 squared plus 𝐵 squared.
Looking down at the right triangle
involving speeds, where we want to solve for the speed 𝑠, we can therefore see that
𝑠 squared equals two squared plus three squared, where here we’re leaving off the
units of miles per hour, which tells us that 𝑠 equals the square root of two
squared plus three squared, which is the square root of four plus nine, or simply
the square root of 13. So this man then was traveling at
the square root of 13 miles per hour as he crossed the river.
And knowing this, let’s now recall
that we want to solve for the total distance he traveled, we call that 𝑑 sub t, as
well as the distance down shore he landed, we call that 𝑑 sub s. Let’s first work on solving for 𝑑
sub 𝑡. And to do it, we can recognize that
this right triangle here with distances is similar to this right triangle here with
speeds. The reason this is so is because
all of these movements and all of these speeds occur over the same time
interval. It’s that amount of time it takes
the man to swim from the south to the north bank. And this means that if we take the
ratio of two of the sides of our distance triangle, say the ratio of 𝑑 sub 𝑡 to
one-half a mile, then this ratio is equal to the corresponding ratio in our speed
triangle. That would be the ratio of the
speeds of the corresponding sides, this speed to this one.
So we write that as a fraction
without the units. And as we mentioned, because these
two triangles are similar, this equality holds. If we multiply each side of this
equation by the distance of one-half mile, that distance cancels on the left. And we find that 𝑑 sub t equals
one-sixth times the square root of 13 miles. This is the total distance the man
traveled in crossing the river.
We record that answer off to the
side. And now we want to solve for 𝑑 sub
s. Recall that this is the distance
downstream that the man lands on the opposite shore. And once again, we can use the fact
that these two triangles are similar. We can say that 𝑑 sub s divided by
one-half mile, that is, the ratio of this side length to this side length, is equal
on our speed triangle to the ratio of this side length to this one. Once more leaving out the units,
that comes out to two-thirds.
Taking this equation, if we again
multiply both sides by one-half mile, we find that 𝑑 sub s simplifies to one-third
of a mile. This is the distance downstream
from where he started that the man lands on the north shore.
Let’s finish up our lesson now by
reviewing a few key points. In talking about relative
velocities, we’ve been reminded that velocities are vector quantities, meaning that
their direction must be taken into account. In solving for relative velocities,
a sign convention is therefore important to establish, whether we do it ourselves or
it’s done for us in a problem statement. And lastly, we saw that relative
velocities can be calculated algebraically or graphically.