Introduction to motion (part 2)

More on how velocity, distance, acceleration and time relate to each other.

All right.

Where I left off in the last
presentation I was dropping a

penny from the top of a
building-- once again, you

should not do, because you
can kill somebody.

Here's the building, and here's
the bad person who's

going to drop something.

Let's say they just hold it
out, and the penny drops.

The penny is going to accelerate
at the rate of

gravity, so it's going to
accelerate downwards at 10

meters per second squared.

Let's start with an interesting
question.

After after two seconds-- and
lets say they drop it right at

t equals 0-- so after two
seconds how fast is it going?

Time is equal to two seconds--
we could even say this change

in time, but we're assuming that
we're starting at time

equals 0, so and change in
time the same thing.

If time is equal to two seconds,
how fast is it going

to be going after two seconds?

Let's use that formula.

Velocity is equal to
acceleration-- acceleration is

the acceleration of gravity,
and that's 10 meters per

second squared-- so velocity
will be 10 meters per second

squared times time, which
is times two seconds.

We can multiply the numbers,
and you get 20.

Just like the numbers, you can
treat the units almost like

variables-- the seconds is the
same thing as this s, so this

s is going to the numerator, and
then you have an s squared

in the denominator.

This s will cancel out with one
of the two s's that are

multiplied down here, so we'll
end up-- actually, let me

write it down-- it'll be 10
meter seconds per second

squared, and that's the
same thing as 20.

That cancels out, this makes
that 1, and so that equals 20

meters per second.

Hopefully, you're starting to
get a little intuition of why

acceleration's units are meters
per second squared.

After two seconds, we're going
20 meters per second.

Let me ask you a slightly more
difficult problem that might

have not been obvious to you.

After two seconds, how far
has the penny gone?

This is interesting.

We have this formula here:
distance is equal to velocity

times time, but the velocity is
changing the entire time.

We know after two seconds that
the velocity is 20 meters per

second, so we could call this
the final velocity-- we'll

called v sub f.

That's just a fancy way of
saying final velocity.

Right when we start
at t equals 0,

what was the velocity?

Right when it started, the
initial velocity-- v sub i,

for initial-- is equal to
0 meters per second.

Can we use this formula?

You might think of a way
to already do it.

Since the acceleration is
constant, and you can only do

this when the acceleration is
constant-- most of what you'll

encounter in a first year
physics course, the

acceleration will be constant,
and especially when you're

dealing with gravity, the
acceleration will be

constant-- you can actually take
the average velocity to

figure out the distance.

So what was the average velocity
over the two seconds?

My final velocity was 20 meters
per second, and my

initial velocity was 0
meters per second.

Obviously, I went continuously
over those two seconds from 0

to 20, so my average velocity--
actually, I've

never seen it done this way
before, but let's just call it

average velocity-- is equal to
the final velocity plus the

initial velocity divided by 2.

I just took the average of the
initial and the final, which

is 20 plus 0-- which is 20--
divided by 2, which is equal

to 10 meters per second.

Right when I let go of the rock,
the ball, or the penny,

whatever I'm dropping, the thing
is stationary, and so

it's 0 meters per second.

After two seconds-- we used this
acceleration formula--

after two seconds,
it accelerated to

20 meters per second.

Over the course of those two
seconds, its average velocity

was 10 meters per second.

We can now use that average
velocity in this

formula right here.

The average velocity, distance
equals average velocity times

time-- you can make a mental
footnote, so it's average

velocity times time when the
velocity is changing and

acceleration is constant, which
is most of what you'll

see in most projectile
motion problems.

Now we could say distance is
equal to the average velocity

times time, which equals 10
meters per second times two

seconds-- once again,
the s's cancel out--

so we're at 20 meters.

After two seconds, not only is
my velocity 20 meters per

second down-- once again, if I
said speed, it would just be

20 meters per second-- but my
distance is the ball, or the

rock, assuming no
air resistance,

has dropped 20 meters.

Hopefully, that makes a little
bit of intuition for you.

If you are taking physics--
which you don't have to view

these videos, that's the idea--
I wanted to show you

that this is actually exactly
like one of the formulas that

you'll see in your
physics class.

It's kind of a shame, but people
tend to just memorize

it in physics without-- when
they're learning projectile

motion without really
appreciating that it just

comes from distance is equal
to velocity times time.

Before, I said velocity
is equal to

acceleration times to time.

Let me just expand that a little
bit, because I assume

that my initial velocity is 0.

Let me just say that the final
velocity is equal to the

initial velocity, because you
could already be going 10

meters per second, and then
you're going to accelerate.

Final velocity is equal to the
initial velocity-- this is an

i-- plus acceleration
times time.

We said that the distance-- we
could rewrite this as the

distance is equal to the average
velocity times time.

I just realized how funny
that character looks.

So, the final velocity is equal
to the initial velocity

plus acceleration times time,
and the distance is equal to

the average velocity
times time.

Let's see if we can use these
two formulas, which we

essentially just applied in
the previous example-- we

didn't do it exactly so
formally-- to come up with a

formula for distance, given
acceleration and time.

We know that the average
velocity-- oh, I switched

colors-- the average velocity is
equal to the final velocity

plus the initial velocity
divided by 2.

What is the final velocity?

The final velocity is equal to
this: substitute, and so the

initial velocity plus
acceleration times time plus

the initial velocity-- my i's
are getting blurred, they're

not showing up-- these are all
i's for initial velocity.

They look like a 2, but I
think you get the idea--

that's all initial velocity.

All of that is over 2.

The average velocity is equal
to the initial velocity plus

acceleration times time, plus
the initial velocity, all of

that divided by 2.

That just equals 2 times initial
velocity-- that looks

like an i now-- plus
acceleration times time

divided by 2, and that equals
the initial velocity plus

acceleration times time
divided by 2.

This might be intuitive for you,
as well, that the average

velocity is equal to your
initial velocity plus-- this

is essentially the difference
between how much you're

accelerating over that time and
speed is going to be that

divided by 2, because we're
taking the average.

If what I just said confused
you, don't worry about it; you

could just backtrack into
what we said before.

Think about a lot of these
formulas yourself, and plug in

numbers, and I think it'll
start to make more sense.

We figured out that the average
velocity is equal to

the initial velocity plus
acceleration times time.

We can just substitute
that back into

this original equation.

Once again, I realized I'm
running out of time, so I'll

see you shortly.