In this chapter, we are trying to find the velocity or acceleration of an object,
with respect to another object. So for example, there are 2 cars going along a road, and you want
to find the velocity of car A with respect to car B, that sort of thing. You will end up using 2 new
equations. This one is for relative velocity and this one is for relative acceleration. Depending
on the question, each of these equations can be written in 2 ways. So let’s quickly go over that.
If a question says you want to find the velocity of object A with respect to B,
then the equation will be written like this. So VA comes first, then VB, and then it’s A over B.
Now if the question says they want the velocity of object B with respect to A, we write it like this.
In this case, B comes first, then A, and then it’s B over A. It’s important to make sure you
get this right and write it properly depending on what the question is asking for or you will
end up with the wrong answers. The same goes for acceleration.
The other thing you need to keep in mind is breaking down velocity and acceleration into
vector components. In a nutshell, what we will do is break the velocity of object A into components,
and then the velocity of object B into components, and we will end up plugging
those values into the equations I mentioned before. The same goes for acceleration. So
let’s go over 4 examples step by step, and see how we can apply these new equations. By the end,
you should be able to solve most, if not all the problems you face in this chapter.
In this question, we need to determine the velocity of plane B with respect to plane
A. The first step is to establish which sides we will say is going to be positive. We will assume
up and to the right to be positive. Next, we need to break down the velocity of each of the
airplanes into vector components. So looking at plane A, we see that it’s travelling straight
down in the vertical direction. That means it will only have the j component, because the I component
is 0. So its negative 500j. It’s negative because it’s going down and we chose up to be positive.
Now, we will focus on plane B. This plane is going at an angle, which means it will
have both components. I’ve drawn the vector components, so we just need to write them
using trigonometry. Establishing them tail to tail will help out a lot. For the I component,
we have 700sine60 degrees and for the j component, we have 700 cos 60 degrees. Let’s simplify this.
Now that we have both velocities, we have to use our relative velocity equation. The
question says we need to find the velocity of plane B with respect to plane A. So, our
relative velocity equation will look like this. Now, we can plug in VB and VA. Let’s solve for
relative velocity. All we are doing is adding up the same components. So I components to I
components and j components to j components. Now, we will find the magnitude of this vector.
Let’s draw the vector components tail to tail to understand what we’re looking for.
Notice how this creates a right-angle triangle. We are trying to find the hypotenuse of
this triangle, which is the magnitude. Finding just the magnitude isn’t enough,
we also need to find the angle that’s created to show the direction. We can find that by
using tan inverse. So the answer to this question, which is to find the
velocity of plane B with respect to plane A, is the magnitude along with the angle.
Let’s look at another example. In this question, we need to find the velocity and acceleration of
car A with respect to car B. The first step is to establish which sides are positive,
so as before, we will pick up and to the right to be positive. Let’s focus on car B first since
it’s going in a straight line. We will write its velocity and acceleration in cartesian
vector form. For velocity, it’s going to have 1 component, which is 100j because the car is
travelling at 100km/h. Now for the acceleration. The question says it’s decelerating at 250 km/h^2,
so the acceleration component is actually facing down, which means it’s -250j.
Let’s leave those aside and look at car A. The velocity is pretty straight forward since at the
instant we are looking at the car, it only has one component, which is 20km/h straight down, so -20j.
For the acceleration, we need to keep in mind that it’s traveling along a curve. You
will remember from previous chapters that objects traveling along a curve will have
a normal acceleration along with a tangential acceleration. So we need to find both of those.
The tangential acceleration is given to us in the question, which is 300 km/h^2. Now we need to find
the normal acceleration. For that, we can use this equation. We know the velocity which is 20 km/h,
and we know the radius of the curve. Since we are using km/h for the velocity,
we need to make sure we convert 100m to km before plugging it into our equation. Solving gives us
the normal acceleration. It’s important to note that these are not in cartesian vector form,
so we need to write them as vectors. First, tangential acceleration. Since it’s accelerating,
the vector is going straight down, so it’s -300j. Now, remember that normal acceleration always
points towards the center of the curve. In this case, due to the position of the car, the vector
arrow is pointing along the horizontal axes towards the center, so the normal acceleration
is negative 4000i. Negative because it’s pointing to the left, and we chose right to be positive. So
the total acceleration of car A is the addition of both normal and tangential acceleration.
So we now have the velocity and acceleration written as vector components. We can start
applying the relative equations. We will start with velocity.
The question says to determine the relative velocity of car A with respect to car B,
so our equation will look like this. Let’s plug in what we found and solve.
Since there is only the j component, the magnitude is simply 120 km/h straight down.
For acceleration, we will do the same, so let’s write our relative acceleration equation.
Plug in what we know and solve. Let’s draw the vector components and solve for the magnitude.
To find the angle, we will use tan inverse. Those are our answers.
Let’s look at another example. In this question, we need to find speed and angle that the man
must direct his boat so it goes from points A to B. This one is a bit different because
we are given the magnitude of velocity, so we are sort of working backwards. As usual,
we will pick up and to the right to be positive. From the question, we know the river is flowing
at 2 m/s. So we can write the velocity of the river flow in cartesian form as 2i.
Next, we need to calculate the angle from points A to B, that’s the angle the boat will
travel along. So for that, we can use tan inverse since it’s a right angle triangle.
Now, let’s break the velocity of the boat into components. We don’t know
the velocity but we do have the angle. So we can write the sine and cosine components
like this. Let’s simplify it a bit. This represents just the velocity of
the boat in cartesian vector form. So we have the velocity of the river, the velocity of the boat,
and now we need the relative velocity of the boat with respect to the river. That’s given
to us as well, which is 5 m/s but we don’t know the angle. This angle is different from
the angle the boat is traveling at so don’t get confused here. Let’s write it in cartesian form.
Now we can write our relative velocity equation. Plug in what we know.
Now we will equate each of the components, so I components to I components and j components to
j components. We get 2 equations with 2 unknowns. All we need to do is solve for them. You can solve
them any way you like, and you will get the velocity and the angle. Those are our answers.
Let’s look at one last example. This one is similar to the 2nd question we did,
the only difference being that the car is actually on a curve at an angle so
let’s see how we can break it into components. Let’s start off by first
choosing up and to the right to be positive. We will focus on car A first and write down
the velocity in cartesian vector form. Since it’s going straight up, it’s going
to be 40j. For acceleration, the question says it’s traveling at a constant velocity,
so there is no acceleration component. Let’s look at car B now. The velocity can
be broken into 2 components. So we get -30sin 30 degrees, and 30 cos 30degrees. Let’s simplify it.
Now, let’s think about the acceleration. The car is on a curve,
so there will be normal and tangential acceleration. The tangential acceleration
is given to us in the question, which is 2m/s^2. Now, let’s find the normal acceleration. For that,
we can use this equation. The velocity is 30 m/s, and the radius of curvature is given to us,
which is 200 m. Let’s solve. Now we have both accelerations, but remember,
they are not in cartesian vector form. Let’s write them as vectors,
starting with the tangential acceleration. So tangential acceleration will have 2 components.
I’ve drawn the components, let’s place them tail to tail. So we have -2sin30 degrees,
and 2 cos 30 degrees. Let’s simplify these. Now for normal acceleration. Remember,
that it always points towards the center of the curve. So we will also break this acceleration
into components like this. I’ve drawn the components along with the new axes, so you
can see that each component, if we put it tail to tail, it creates a triangle with a 30-degree
angle. So we get negative 4.5cos30 degrees and negative 4.5sin30 degrees. Let’s simplify these.
Now that we broke both accelerations into vector components, the total acceleration of car B is
the addition of all of these components. So we are going to add the components together.
Everything we need is calculated, so we can start applying our relative velocity and acceleration
equations. Let’s start with velocity. The question says we need to find the velocity
of car B with respect to car A. So our equation looks like this: let’s plug in what we found and
solve. Now we will calculate the magnitude. The angle can be found using tan inverse.
Now for acceleration. Let’s use the relative acceleration equation and plug in what we know
and solve. Let’s find the magnitude now. The angle can be found using tan inverse.
Those are our answers. That pretty much covers the problems you will face in this chapter. If this
video helped you, please give a thumbs up. Thanks for watching and best of luck with your studies!