Hi there - in this video I'll talk about
the equations of motion.
I'll show you how to derive all four equations then go over some important points to remember.
I've also made a video of worked examples so take a look at that too.
Sometimes, these are referred to as the suvat equations because of the letters used.
s is displacement in meters, u is initial
velocity in meters per second,
V is final velocity also in meters per second, a is acceleration in meters per second per second
and finally, t is time in seconds.
Now, it's important to remember that the first four quantities are all vectors
so, for them, direction has to be taken into account.
I'll say more about this after
going through the derivations.
Here's the first - we'll start with an equation we already know from National 5: a=(v-u)/t
And rearrange it to give the new equation v=u +at
It's a good idea to practice writing
down these derivations yourself
so, in the first line, you'd write a=(v-u)/t
Eventually, we want to get v to be the subject of the equation but we can't do that all in one step.
We'll start here. So, in the second line we have v-u=at
If we add u to each side, we get v=at+u
so we can just rearrange the right-hand side
to give us the equation we're looking for.
As I said, try writing out derivations with each step on a new line until you've memorised them.
Here's the second one. This one involves a velocity time graph and something else we know from National 5 -
that displacement can be calculated by finding the area under the graph. We'll
use this to derive the equation s equals
We'll use this to derive the equation s=ut+1/2 at^2
Imagine an object accelerating uniformly from initial velocity u to final velocity v
in time t seconds.
The displacement, s, is found from the area under the graph.
So, firstly, we have this section a rectangle.
It's area can be found by multiplying
its height by its length giving us u t
The second section is a triangle. Its area
is equal to 1/2 times the height times the length.
The height of the triangle is
v-u and its length is just t.
So it's area is 1/2 times (v-u) times t.
If we remember back to the last derivation, at one stage we had v-u=at
So we can substitute at into the brackets.
If we now multiply out
the brackets we can get the equation we're after.
For the third derivation we
start with the first equation and end with this - v^2=u^2=2as
We start by squaring both sides which gives us this. We then want to multiply out the brackets.
There's more than one way of doing this but we should end up with v^2=u^2+2uat+a^2t^2
Obviously, you'll want to try this yourself.
The left hand side of the
equation looks fine but this part - it needs more work.
What we do is take out 2a from this part of the equation, leaving the rest in brackets.
Dividing 2uat by 2a  gives ut. If we divide a^2 t^2 by a, we would get at^2.
So dividing by 2a would give 1/2 at^2
So the right hand side of the
equation can be written like this.
Go over this step again if you're not
too sure.
Now, remember that our second equation was s=ut+1/2at^2
So the term in brackets can be replaced by the letter s, giving us our third equation.
The last derivation is the simplest. We start with these two equations.
The first says that displacement is equal to average velocity times time
and the second is an equation for average velocity which is 1/2 (u+v).
We then end up with this: s=1/2(u+v)t
All we have to do substitute the second equation into this position in the first like so
To end the video, I'll go over a few
final points by answering a question
Remember before that I said that s,u,v and a were all vectors and that only t was a scalar.
That's very important when
answering questions like this one.
So the question says a girl throws a ball
vertically upwards with an initial velocity of 20ms-1
then catches it sometime later.
Calculate the time taken for the ball to return to the
same height as it was released.
Now, in this question, I'm going to use something known as sign convention.
So I'm going to be taking as positive the upwards direction.
Now, I've written that here in shorthand but really what I should be writing at the start of this answer is something like this.
Taking upwards as positive.
Now, if I then write down what I know and, of course, what I'm trying to work out.
Remember we said that these were known as the suvat equations.
Now, s - that's displacement not distance.
If it was distance, we don't actually know
how far up the ball has moved to maximum height.
And then, of course, the total
distance by the time it comes back to its starting position,
its total distance would be twice the maximum height.
This is displacement remember and,
of course, if it's returned to its initial position
then of course its displacement will be zero.
Now we've said, using this sign convention that anything
moving upwards is positive
so initial velocity would be +20ms-1 and of course that then means
that final velocity, v (since the ball is
moving downwards) when it's at the same height
then of course we're assuming at
this point that the air resistance is negligible.
It has no effect so of course,
coming back to the same height, final velocity would be -20ms-1
Acceleration due to gravity - that's 9.8, which you'll find in the data sheets at the start of the paper...
but of course when the ball is moving upwards in the positive direction, it's decelerating.
That would give a negative acceleration. Then, when it starts to move downwards,
it's accelerating so it's accelerating, but that's in the negative direction
So, it's either decelerating in the positive direction or accelerating in the negative direction.
Either way, that gives us an acceleration of negative 9.8 ms^-2 when we're, of course, taking upwards as positive.
Now, time we don't know. And it's a simple case of just working out which equation to use
I'm going to use v=u+at and then, of course, substitute.
V is -20, is equal to 20+(-9.8t)
What I'm doing next is, I'm going to subtract 20 from both sides
I'm also going to switch
the right hand side with the left hand.
And what that will give me is
-9.8t=-40
And of course, to work out t, I would divide both sides by -9.8.
So of course at that stage, t is equal to -40 divided by -9.8 is equal to, and we'll get the calculator.
-40 / -9.8 = (4.081632653). That's a big number!
Now if we look at the question, of course, the one number we are given in the question is 20.0ms^-1
That's the initial velocity, that's three significant figures so of course we should be writing this to the
same number of significant figures. That
would, of course, then mean that time is 4.08 seconds.
Now if you've done that question yourself and you've wrongly applied the sign convention...
(that's where people start getting a negative time). That's where you know that you've done something wrong.
OK, so if you've tried this question
yourself you should be getting this answer: 4.08
Now remember to look out for the video of example questions.
Getting to grips with the equations of motion is very important since you'll be using them to answer projectile questions
For more information on upcoming videos, summary sheets and so on visit physics-podcast.co.uk
Thank you for listening!
