Kinematics is a branch of classical mechanics that describes the motion of points, bodies or any other objects, without considering the mass of each of the forces that caused the motion.
There are two distinct types of motion.
- One-dimensional motion.
- Two-dimensional motion.
It involves anything that moves along a straight line. Velocity, acceleration, and time are defined before. They used in four equations for motion like this. We know already.
Here the acceleration is constant in all the problems. Free-falling bodies are examined as an example of one-dimensional motion.
Here, as an example, we can consider the projectile motion of a body we use the above equations, with the addition of vectors and components. Let’s we learn more about this later. Now we continue with relative motion.
In this lesson, A certain frame can be defined in the form of either a basic procedure or a frame structure. It can be considered as the frame where the observer is taking the reading for the experiment. Think for a moment,
I am the observer. When I’m standing on the road we can see vehicles are moving. I’m standing on the road, but I’m(c) not moving as shown in fig. When I in the place of “A” the certain frame will be the A and its surroundings. At that time, I am the observer. When I in the place of “B” the certain frame will be the B and its surroundings.
Ok, now you know what is a certain frame. Now we going to talk about relative motion.
Relative motion is the calculation of the motion of an object with regard to some other moving object. According to the calculations of relative motions, we can be divided them into three fields as the following.
- Motion in the same direction.
- Motion in the opposite directions.
- Motion when a θ angle is formed between directions of A and B.
Motion in the same direction
We consider two objects A and B moving in the same direction in parallel paths. The velocities of A and B are respectively VA and VB. lets us obtain the velocity of A at the sight of an observer at B. Here,
The velocity of A relative to B = (x-y) its direction is to the → left to right as shown in fig. If x>y the direction of (x-y) is the same → left to right. If x<y the direction of (x-y) is the same ← left to right because the (x-y) <0. If (x-y) =0 Observer can see the A is not moving. Now we understand the resulting velocity of A relative to B is (x-y) when A and B are same directions.
Motion in the opposite directions.
A and B moving in the opposite direction in parallel paths. The velocities of A and B are respectively VA and VB same as the above. Lets we obtain the velocity of A at the sight of an observer at B. Here, the velocity of A relative to B is always (x+y) because of the position of the opposite. There is no zero value for the (x+y). You want to understand how is the equation shown in fig. AB=AE+EB. We can replace E with any certain frame.
- A boat has a speed of 3.0ms-1 in still water. It is used to cross a river 1000m wide along. The water flows speed at 1.8ms-1 in the river. The boat is directed towards the opposite bank but the water carries it downstream. Calculate
- the speed of boat relative to the bank of the river,
- the distance the boat is carried downstream from its point of departure to its point of arrival at the opposite bank,
- the time is taken to cross the river.
After reading the problem we drow the sketch for it as shown in this fig. Let we assume
B = Boat, W = Water and E = River Bank is meaning Earth. In the question, you can see
“A boat has a speed of 3.0ms-1 in still water”.
It’s mean BW = ↓3.0ms-1 . “The water flows speed at 1.8ms-1 “It means WE =↓1.8ms-1. That’s only we can drag from the problem which helps us to solve it.
A. In part A we want to find BE. Next, Draw the velocity triangle of BE = BW + WE
direction = angle between BE and the bank
Motion when a θ angle is formed between directions of A and B.
Now lets we learn how is this position. There many steps to solve any relative motion problems. So we want to follow them. Following are the steps.
- First, we want to understand the problem. To do that we should read carefully without any doubt.
- Sketch down the problem graphically. It should be understandable. It should help you to solve the problem correctly. It should be contained with all data of the problem.
- Next we drow the velocity triangle for the problem.
- Now we get the equation that we want to solve it.
- Applying the values to the equation and get the answer.
We can solve any problem of relative motions by the above 5 steps.
For A, B, C’s three Certain frames,
(AB) = (AC) + (CB) .
The speed of A and B are respectively
VAE = y and VBE = x.
The angle of y and x is θ. Now we want to find the velocity of A relative to B.
It is very easy,
you can see it is done in this fig.
- Two ships, A and B, leave a port at the same time. Ship A travels due north at a steady speed of 10kmh-1 and ship B travels N600E at a steady speed of 15kmh-1.
- What is the distance and direction from A to B after 1 hour?
- What is the velocity of B relative to A?
- What is the direction of
First, It is necessary to know about trigonometry rules for quick the solution. This is the cosine rule of trigonometry. In this fig, there is a rhythm between a, b, c, A, B, and C. If we understood it we can remember it well. By this, you can save time.
- After one hour, ship A has moved 10km to the north and ship B has moved 15km to the N600E. N600E mean, 600 is measured from North to East.
distance and direction from A to B after 1 hour = AB (as shown in fig)
Now use above formula,
- To find the velocity of B relative to A we want to
draw the velocity triangle. In this fig, now we can say what are the values of the above formula.
- Next, we can find the value of α
α = angle between the North and velocity of B relative to A.