Without movement this universe would be non-existent. Even if the Earth was present, there would be no life because movement is an essential part of biology. From the invisible interactions of atoms to rotations of large black holes and galaxies, movement plays an important role. Therefore, in our mission to better understand this complex universe through physics we should start by looking at Kinematics.
Kinematics is the study of movement. In this section we will learn about the different kinematic equations and how to use them to determine the velocity or position of an object in a two dimensional space. Before we do that, let's remind ourselves of the basics.
Speed tells us how fast an object is moving. In other words it is the rate of change of distance. It does not ,however, tell us the direction in which the object is moving. Which is why speed is a scalar measurement. Knowing just the speed without direction is not very useful. Therefore, expressing speed as a vector quantity is essential.
The first step in converting speed into a vector quantity is understanding the displacement of the object being observed.
Displacement is the vector quantity you get when distance (scalar quantity) is expressed with direction. What does this mean? Well distance tells us how far an object has moved by considering the path it takes. Whereas, displacement tells us the direct distance of that object between the start point and end point. This is illustrated in the image below.
Now how can we express direction? Well we can use a positive sign to show the object is moving in one direction and a negative sign to show that it is moving in the opposite of that direction. So, if we were to assume that the start point is the origin along the x-y plane then the end point is in the positive quadrant; therefore, the displacement is positive. However, if the end point was below the start point it would be in the negative y-quadrant which means the displacement is negative. You can also express direction using cardinal directions or trigonometry as we will see later on.
Let's look at an example.
Let's say you do a round trip to Colombo from kandy ( Colombo to Kandy ~ 123 km) in 2 hours. The distance you traveled is 123 km + 123 km = 246 km. But the displacement is 0 km. You traveled away from Kandy for +123 km and traveled back to Kandy for -123 km which means you are back at the start .
So we already know the formula of speed. We need to find a way to include direction into speed for it to be converted to velocity. To do this we can replace distance with displacement to get the equation shown below. From this equation we can say that velocity is basically the rate of change of displacement. Make sure not to confuse the s in this equation with speed because s stands for displacement.
But is velocity really different from speed? Yes, they are different. For example if a person was travelling north at 100 km/hr and then travels south at 100 km/hr; then his speed in both cases is the same. However, if we were to look at the velocity, then they are different. When he is travelling north then his velocity would be +100 km/hr however, on his way south his velocity will be -100 km/hr. As you can see even though magnitude of speed and velocity is the same, because of direction they are different.
Average Speed and Instantaneous Speed/Velocity
Objects that are moving don't really have a fixed speed in reality. The speed at one point in time will probably be different at another point in time. Let's look at Example 1 again. When you are travelling in a vehicle, it doesn't move at fixed speed. You have to break from time to time and stop at traffic lights and so on. So then, what is that speed that we calculated? Why is it a fixed value? Well, that is the average speed for the journey. It is what you get by diving the total distance travelled by the total time taken.
Then, what is instantaneous speed? This is the speed of the object at a particular point in time. Average speed is basically the average of all the instantaneous speeds for the duration of the object's motion. For example, the speed shown in the vehicle odometer is the instantaneous speed.
When comparing instantaneous speed and instantaneus velocity, we will always observe that even though the direction could be different the magnitude is always the same. However, If we were to look at average speed and average velocity then the changes could be more pronounced depending on the situation. Let's look at Example 1 again:
As you can see the values are totally different. The magnitude is different. Always be careful to distinguish between average data and instantaneous data.
So we know that when an object moves, it covers a distance. As it keeps moving the distance it covers keeps changing. To explain this, we can use speed or velocity because it tells us the rate of change of distance or displacement. Then what about velocity? Velocity also changes as the object keeps in motion. To measure the rate of change of velocity, we use acceleration:
Since velocity is a vector, so is acceleration.
Graphs can help us see an object's movement in clear detail because all of the data is visually represented. Let's look at distance-time graphs, displacement-time graphs and velocity-time graphs.
Let's look at Ed's trip to work and back. We'll express time in hours. At 0 hours he leaves home. After 30 minutes he has travelled 10 km and makes a stop at his favorite donut shop to get breakfast. He stays there for 30 minutes. After that he leaves and drives for another 30 minutes traveling for 10 km to reach to his office. He works for 4 hours. Then he has to go to a lunch meeting with a client who is 20 km away from work (40 km away from home). It takes him an hour. The meeting takes two hours and finally he drives back home directly without stopping at work. It takes him 3 hours. He drives past home for another 30 minutes to make a stop at his parent's house which is 15 km away from home. He spends an hour there and then finally drives back home. The final trip takes 30 minutes.
Plot distance/displacement vs time and speed/velocity vs time graphs for Ed's trip.
Let's plot a distance- time graph first:
Now let's plot the displacement-time graph. For this graph let's use home as the starting point, which means Ed's location is always relative to his home.
As you can see from the two graphs, displacement and distance are different.
In the displacement-time graph, the zero on the y-axis represents his house's location. So when he travels away from his house to work, it's positive. But when he travels away from his house to his parents house, opposite direction, it's negative because its in the opposite direction. It is important to note that in this simple example we are assuming that he travels from home to work and parents house in a straight line.
If you compare both graphs you will notice that the gradients are similar. When they are different, it's because they have opposite signs, the magnitude is the same. What can you deduce from this? Well, the gradients of these graphs are equal to the speeds or velocities of the object at different points in time.
Now lets plot the speed-time and velocity-time graphs:
Similar to the distance-time graph, speed-time graph does not break the origin line; its always on the positive side of the quadrant. However, the velocity-time graph breaks through the origin into the negative side. The same happened with the displacement-time graph.
From the velocity/speed - time graphs, the distance or displacement of the object can be calculated by calculating the area under the graph. The area must always be calculated by using the x-axis as the base. Calculating the total area under the graph will give you the total distance travelled or total displacement. We know that the gradient of the distance/displacement - time graphs gives us the speed/velocity. But what does the gradient of the speed/velocity - time graphs tell us.
This gradient is equal to the acceleration of the object. Now in this example we didn't account for acceleration or deceleration. We assumed that he reached the speed he was travelling at instantly and he also stopped instantly. What I mean by instantly is that it took him 0 amount of time to reach his top speed from 0 m/s and vice versa This can be seen in the graph as well because the gradient are undefined.
The above graphs are very simple representations. Usually these graphs are curves and lines with varying gradients. But whatever the graph shape maybe, the concepts are the same. Now that we understand the basics let's delve in deeper.
Now, we know that displacement = velocity x time. This is off course for a situation where acceleration is zero. This makes our calculations easy since velocity is uniform. However, in some cases objects do accelerate and in those cases we can't use the above equations. For such situations we will need the kinematic equations that we will now derive.
Before we proceed we will have to make an assumption : The acceleration of the object is always uniform. This assumption is a must if the kinematic equations are to work.
The variables we will be using are shown below:
This is the equation for acceleration that we discussed in the previous section. Let's rewrite this by using the variables above.
Change in velocity can be replaced by u and v since they are initial and final velocities. The new equation we get is:
We can rearrange this equation to make final velocity the subject of the equation. This is imprtant becasue we are going to use it to derive another equation later on.
Kinematic Equation 1
As discussed earlier, displacement = velocity x time. But since we have acceleration, the velocity is not constant. Therefore we are going to have to use an average velocity for the equation to be correct in a case where the object is accelerating. So average velocity will be:
If we sub this equation into our original displacement equation we get the following:
Kinematic Equation 2
We can use the two equations above to derive another two:
Kinematic Equation 3
Kinematic Equation 4
Okay so we have now derived our kinematic equations. Let's look at how we can actually apply them.
These equations can be used to determine the motion of a projectile. A projectile is an object that has been thrown by applying a force on it. Usually the main force acting on a projectile is gravity. However, there are various other external forces like air resistance that act on them; but we will only focus on gravity right now.
If we are using these equation in 1-dimension then it's straight forward. You just plug in the values for the known variables to calculate the unknowns. However, projectiles are not occurring in 1 dimension. Projectile moves in both the x and y axis. We have to be very careful in these cases since most of the variables in the kinematic equations are vectors. Therefore, we have to account for direction before plugging in the values.
Before dealing with projectiles let's try to figure out how to separate vectors into vertical and horizontal components. Look at the image below.
The object is moving at 15 m/s in a path that is 35 degrees from the x-axis. For this object to be able to move to the right while at the same time moving up it must have speed in both the x and y direction. To find the velocity in the x direction we have to multiply the given velocity by cos (35). For the velocity in the y direction we have to multiply it by sin (35) as shown below.
Looking at the values we calculated, it is clear that the object is not moving at the same speed in both x and y direction. What does this mean? Well it means that it is covering distance at a faster rate in the x direction than it is in the y direction.
Which way would the object have to move for both the x and y components to be the same? It would have to adjust its path to 45 degrees from the x-axis as shown below:
By determining the components of the velocity we are now able to precisely change the object's path to go exactly where we want the it to go. For example if there are things blocking its pathway we now know how to change the velocity in order to avoid those hurdles. Therefore, understanding how to split vectors into their respective components is very important.
Alright let's look at how we can use this knowledge to tackle projectiles. We will do this while tackling a simple problem.
Assume someone is standing on the top of a wall as shown below. He throws a ball at 2 m/s aimed 60 degrees above the x-axis. The height between the ground and ball is 60 meters just before he throws it. The ball reaches the ground and does not rebound. Calculate:
i) The total time the ball is in the air,
ii)How far away from the wall the ball lands.
Note that the diagram is not drawn to scale.
Okay, so where do we start? First let's recall the 4 kinematic equations we derived earlier:
Let's first calculate the x and y components of velocity:
Now we are going to make an assumption that the x-component of velocity will be fixed because there is no resistance to motion in the x-direction. We are assuming that air resistance is negligible. Therefore, only the y-component will change because gravity is acting on the object along the y-axis.
This means that the ball will move up due to the initial velocity. The y-component of velocity will turn to zero because gravity is initially acting against the ball causing it to decelerate. Then the velocity will increase as it free falls because gravity is now causing it to accelerate. The acceleration or deceleration due to gravity is always -9.81 m/s^2 and it is denoted by g. g is always negative because gravity is always acting downwards.
Let's look back at the diagram above. We can assume that the x-component of velocity will not change through out it's motion because gravity only acts on the y-axis. This means that whatever distance, x meters, it travels will only depend on Vx and time it is airborne. So then which equation can we use to find x. We know that initial velocity and final velocity is the same for the x component of velocity. We also know that the ball will move in the x-direction for as long as it is airborne. Therefore, if we can find total time (t) the ball is in the air, we can use the Equation 1 to find it.
To find t, we will have to break the motion into 2 parts. First part is the time the ball takes moving up against gravity due to the initial Vy. The second part is the time it takes for the ball to fall to the ground from the highest point it reaches. We will have to find how high it moves to determine the total distance the ball has to fall. The calculation for both parts is shown below:
First we calculate the time it takes the ball to reach maximum height from the point of release. Then we can use this time, t1, to find how high the ball travels. Then, we move to the second part.
Now that we calculated the times for both parts, we can add them to get the total time the ball is airborne:
We can now at last calculate how far away from the wall the ball lands: