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# Kinematic Equations (E01: introduction)

Hi everyone, in this video we'll be looking at the kinematic, or SUVAT equations, starting
with what they are and how they are used, and ending with how they can be derived. There
will be 2 more episodes after this, showing how to solve slightly more complex problems,
such as calculating how long it will take for an object to catch up to a moving target;
and calculating the velocity required to shoot a ball through a hoop.
For now, consider this red dot as it moves from point A to B. There are five main things
we can list about its motion: It's overall change in position — in other words, displacement,
which let's pretend is 8 metres to the right; then there's the time it takes to travel between
the two points, which here is 2 seconds; the object's initial velocity, which here is 1
metre per second; its final velocity, which is 7 metres per second; and its acceleration,
which is 3 metres per seconds per second.
When writing the equations, displacement is usually denoted by the letter s; initial velocity
by 'u', final velocity by 'v', acceleration by 'a', and time by 't'. As you can see, these
variables form the acronym SUVAT.
So each of the suvat equations uses a different combination of four out of these five variables.
Thus, as long as we know the values for any three of them, we can use the equations to
calculate the unknowns. It's important to note that these equations are only applicable
to objects moving with constant acceleration, and in a straight line.
Let's do an example. Say we're making a video game, and we want our character to be able
to jump a certain height. The first thing we need to do is decide which direction is
positive. I'm going to say that upwards motion is positive, and downwards motion is negative.
Thus if we want our character be able to jump, say, three metres high; then the displacement
from the ground to the apex of the jump will be positive 3 metres. We also need to choose
a value for the acceleration due to gravity in this imaginary game world. I'll pick 20
metres per second squared, and this is downwards, so that value will be negative.
The question is now, what must the initial velocity of the character be in order to reach
a height of 3 metres. Remember that we need three pieces of information in order to calculate
the unknown, so we need to find one more known value. As it happens, we know the final velocity
here, since at the very peak of the jump the character will be moving neither up nor down,
and so we can say that the final velocity is zero. Now we just need to find the equation
that uses the four variables we're interested in, which are (s,u,v, and a). As you can see,
the only equation that has this combination is number 5. We're trying to find initial
velocity, so lets rearrange this with initial velocity on the left hand side. We can then
substitute in the values we know, to get u squared = 120. We then take the square root
of both sides, to get u = 10.95. Now that we now the initial velocity, we can make the
character jump the height we wanted.
So a lot of the time it's really going to be that simple. You'll have three known values,
and you just plug them into an equation to find the fourth. For the rest of this video
we're going to be looking at how these equations are derived.
Speed is a measure of how far an object travels in a given amount of time. You're no doubt
familiar with the equation "speed equals distance over time". Since velocity is just speed with
a direction, we change the wording to "velocity equals displacement over time". And of course,
if we consider that the object may be accelerating, then we must say "average velocity equals
displacement over time". The average of two numbers is just both of them added together,
and then divided by 2. So, average velocity - or, (vi + vf)/2 - is equal to d/t.
If we substitute in our variable names, this is actually one of our kinematic equations,
although its usually re-arranged by multiplying both sides by time, to get it in the form:
d = (vi+vf)/2 * t.
Acceleration is a measure of how much velocity changes over time. Since we're assuming constant
acceleration, we can say that acceleration is simply equal to the change in velocity
over time. Which we can rewrite as acceleration = (final velocity - initial velocity), over
time. If we substitute in our variable names, we have our second kinematic equation, although
usually it's re-arranged, by multiplying both sides by time, and then adding initial velocity
to both sides, to get final velocity = initial velocity + acceleration * time.
The remaining three equations are derived by combining these first two in different
ways.
So, for example, we can replace the final velocity in the first equation with what we
calculated final velocity to be in the second. The two U's add together to make 2u, and then
if we perform the multiplication by t we get 2ut + at^2, all over two. Simplifying this
leaves us with displacement = initial velocity * time + acceleration*time squared/2. That's
the third equation.
Lets now rearrange the second equation to solve for initial velocity. We substitute
that into the first, and simplify in the same way as the previous one, this time to get
displacement = final velocity * time - acceleration * time squared/2. This is the fourth equation.
For the final equation, we will rearrange the second equation to solve for time, and
plug that into the first equation. Note that the top part is a difference of squares, so
when we multiply out we'll get s = v^2 - u^2/2a. This is usually rearranged by multiplying
both sides by 2a, and then adding u^2 to both sides to end up with: final velocity squared
= initial velocity squared + 2*acceleration * displacement. Which, of course, is what
we used to solve the jumping problem.
Alright, I hope you found this video helpful, make sure to check out the following two episodes
for some more complex problems. Until then, cheers.