What I want to do with this video
is think about what happens to some type of projectile,
maybe a ball or rock, if I were to throw it
straight up into the air.
To do that I want to plot distance relative to time.
There are a few things I am going to tell you about
my throwing the rock into the air.
The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s)
I picked this initial velocity because
it will make the math a little bit easier.
We also know the acceleration near the surface of the earth.
We know the force of gravity
near the surface of the earth is the mass of the object times the acceleration.
(let me write this down)
The force of gravity is going to be the mass of the object times little g.
little g is gravity near the surface of the earth
g is 9.8 meters per second squared (9.8m/s^2)
Now if you want the acceleration on earth
you just take the force divided by the mass
Because we have the general equation
Force equals mass times acceleration (F=ma)
If you want acceleration divide both sides by mass
so you get force over mass
So, lets just divide this by mass
If you divide both sides by mass,
on the left hand side you will get acceleration
and on the right hand side you will get the quantity little g.
The whole reason why I did this is
when we look at the g it really comes from
the universal law of gravitation.
You can really view g as
measuring the gravitational field strength near the surface of the earth.
Then that helps us figure out the force
when you multiply mass times g.
Then you use F=ma, the second law,
to come up with g again
which is actually the acceleration.
This is accelerating you towards the center of the earth.
The other thing I want to make clear:
when you talk about the Force of gravity
generally the force of gravity is equal to big G
Big G (which is different than little g) times
the product of the masses of the two things
over the square of the distance between the two things.
You might be saying "Wait, clearly the force of gravity is dependent on the distance.
So if I were to throw something up into the air,
won't the distance change."
And you would be right!
That is technically right, but
the reality is that when you
throw something up into the air
that change in distance is so small
relative to the distance between the object and the center of the earth
that to make the math simple,
When we are at or near the surface of the earth (including in our atmosphere)
we can assume that it is constant.
Remember that little g over there is
all of these terms combined.
If we assume that mass one (m1) is
the mass of the earth, and
r is the radius of the earth (the distance from the center of the earth)
So you would be correct in thinking that it changes a little bit.
The force of gravity changes a little bit, but
for the sake of throwing things up into our atmosphere
we can assume that it is constant.
And if we were to calculate it
it is 9.8 meters per second squared
and I have rounded here to the nearest tenth.
I want to be clear these are vector quantities.
When we start throwing things up into the air
the convention is
if something is moving up it is given a positive value,
and if it is moving down we give it a negative value.
Well, for an object that is in free fall
gravity would be accelerating it downwards, or
the force of gravity is downwards.
So, little g over here,
if you want to give it its direction,
is negative. Little g is -9.8m/s2.
So, we have the acceleration due to gravity.
The acceleration due to gravity (ag) is
negative 9.8 meters per second squared (9.8m/s^2).
Now I want to plot distance relative to time.
Let's think about how we can set up a formula,
derive a formula that, if we input time as a variable,
we can get distance.
We can assume these values right over here.
Well actually I want to plot displacement over time because that will be more interesting.
We know that displacement is
the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)).
Right now we have
something in terms of time, distance, and average velocity
but not in terms of initial velocity and acceleration.
We know that average velocity is the same thing as
initial velocity (vi) plus final velocity (vf) over 2. (Vavg=(vi+vf)/2)
If we assume constant acceleration.
We can only calculate Vavg this way assuming constant acceleration.
Once again when were are dealing
with objects not too far from the center of the earth
we can make that assumption.
Assuming that we have a constant acceleration
Once again we don't have what our final velocity is.
So, we need to think about this a little more.
We can express our final velocity in terms of our initial velocity and time.
Just dealing with this part, the average velocity.
So we can rewrite this expression
as the initial velocity plus something over 2.
and what is final velocity?
Well the final velocity is going to be
your initial velocity plus your acceleration times change in time.
If you are starting at 10m/s
and you are accelerated at 1m/s^2
then after 1 second you will be going 1 second faster than that. (11m/s)
So this right here is your final velocity.
Let me make sure that these are all vector quantities...(draws vector arrows)
All of these are vector quantities.
Hopefully it is ingrained in you that these are all vector quantities, direction matters.
And let's see how we can simplify this
Well these two terms
(remember we are just dealing with the average velocity here)
These two terms if you combine them become 2 times initial velocity (2vi).
two times my initial velocity
and then divided by this 2
plus all of this business divided by this 2.
which is my acceleration times my change in time divided by 2.
All of this was another way to write average velocity.
the whole reason why I did this is because we don't have final velocity
but we have acceleration
and we are going to use change in time as our independent variable.
We still have to multiply this by this green change in time here.
multiply all of this times the green change in time.
All of this is what displacement is going to be.
This is displacement, and lets see...
we can multiply the change in time times all this
actually these 2s cancel out
and we get (continued over here)
We get: displacement is equal to
initial velocity times
change in time
Some physics classes or textbooks put time there but it is really change in time.
change in time is a little more accurate
plus 1/2 (which is the same as dividing by 2)
plus one half times the acceleration
times the acceleration
times (we have a delta t times delta t)
change in time times change in time
the triangle is delta and it just means "change in"
so change in time times change in time
is just change in times squared.
In some classes you will see this written as
d is equal to vi times t plus 1/2 a t squared
this is the same exact thing
they are just using d for displacement
and t in place of delta t.
The one thing I want you to realize with this video
is that this is a very straight forward thing to derive.
Maybe if you were under time pressure you would want to be able to whip this out,
but the important thing, so you remember how to do this when you are 30 or 40 or 50
or when you are an engineer and you are trying
to send a rocket into space and you don't have a physics book to look it up,
is that it comes from the simple displacement is equal to average velocity times change in time
and we assume constant acceleration,
and you can just derive the rest of this.
I am going to leave you there in this video.
Let me erase this part right over here.
We are going to leave it right over here.
In the next video we are going to use this
formula we just derived.
We are going to use this to actually
plot the displacement vs time
because that is interesting and we are going to be thinking about
what happens to the velocity and the acceleration
as we move further and further in time.