Acceleration“If the change in position over time is

velocity, what is the change in velocity over time?”

Goals: learn how acceleration effects velocity and position.

Derive an equation that describes final position under constant acceleration.

Find a way to solve for time in the equation that will be derived.

That is a good question – Let’s review what we already know.

What does it mean to change our position over time?

It simply means that we move from one point to another in a certain amount of time.

Remember, we measure velocity in meters/second, or m/s.

What does velocity over time mean???

Let’s think about this! We know that position over time is measured in m/s and that position is measured in meters.

Maybe if we just use the same idea, we can get the right units for this thing called acceleration.

So, velocity is measured in m/s so if we divide by seconds, we get…

Units of Acceleration! If Velocity = , then dividing by seconds

will look like: Acceleration = or: Acceleration = Why would we want to divide by

seconds twice?!?!!?!?? That makes no sense!!!!!!!!!!!!!!!!

Why would we want to divide by seconds twice?!?!!?!??

Don’t think of this as dividing by seconds twice – we are simply making a complex measurement that happens to have 2 factors of time.

Think of the units as = , not Although, we do abbreviate the units

as: m/s2 .

How do we measure acceleration???

We thought of velocity as the time it took to get from one position to another, so we will think of acceleration as:

The time it takes to get from one velocity to another!

A positive acceleration means that we are speeding up in the positive direction, while a negative acceleration means we are speeding up in the negative direction.

Let’s think about this a little deeper –

We know that a constant velocity changes something's position, but how does acceleration change things…Does it change velocity and position?

YES! It changes both! In a 2 dimensional world, we have 4

possible ways to think about velocity and acceleration: 1 where both are positive, 1 where both are negative, and 2 where they are opposites!

Lets try each one! Acceleration gets a blue arrow,

while velocity get the red arrow Lets talk about the examples where

they are both the same, either both positive or negative

In this example, we are moving forward (positive velocity,) but every second that we are moving, we are moving forward faster! (positive acceleration)

Negative velocity & Negative acceleration:

Now, we are moving in the negative direction (negative velocity,) but every second that we are moving, we are moving in the negative direction faster! (negative acceleration)

This is where it gets tricky!!!

Now, we are moving in the negative direction (negative velocity,) but every second that we are moving, we are speeding up in the positive direction! (positive acceleration)

Whoa, what does that mean??? If we are speeding up in the opposite

direction we are moving, then we slow down (until net velocity is zero,) then we will actually move in the direction of the acceleration – positive direction in this case

This is where it gets tricky!!!

In this example, we are moving forward (positive velocity,) but every second that we are moving, we are speeding up in the negative direction (negative acceleration)

Even though our acceleration and velocity are in opposite directions, if we are speeding up in the opposite direction we are moving, then we slow down (until net velocity is zero,) then we will actually move in the direction of the acceleration – the negative direction in this case!

Wow, that was a lot to take in! Yes it was, but we still need to

accomplish two goals: What equation can we use to find the

final position of a moving object with a constant acceleration?

We already know an equation for moving under constant velocity:

x = xo + v(t) where x is the final position, xo is the final position, v is velocity and t is time.

How did we figure that out again?

Don’t forget how we figured it out!!! We know that velocity is measured in

m/s, so if we want to convert to position (just meters,) all we need to do is multiply by a time (whichever time we want to use) to turn it into a position!

Maybe we should try this again for acceleration?

YES!!!

Let’s try it! We already have our first equation: x = xo + v(t) Now we just need to add in a part that

deals with acceleration! How can we turn acceleration into a

position? Well, we already know how to turn a

velocity into a position, so we just need to find a way to turn acceleration into a velocity!

How do we do that??? Well, it shouldn’t be that hard, we

already know that: Acceleration = , so we just need to

multiply by a time to get convert to velocity – if we do that we get:

x = xo + v(t) + a(t)(t) or x = xo + v(t) + a(t2) Wow, that wasn’t too hard!

Is that really all we have to do? No, unfortunately it is not quite that

easy. It actually took a long time to figure out

why that didn’t work. The reason comes from one of the

most complicated branches of math – Calculus.

Some of you will be lucky (or unlucky, depending on how much you like math,) enough to study Calculus in high school!

What is the actual equation?

In reality, we were actually very close to being right the first time!

It is: x = xo + v(t) + I won’t even try to explain why I’m sure you won’t mind : ) If you would really want to know, as me

about it after we work on the simulations you will be doing.

Wait…Simulations!!! Yes, after we finish with this

powerpoint, you will do some simulations to help you cement your understanding.

First, we still need to figure out how to solve for time in our equation.

This is harder than you might think! Let’s do an example problem on the

next slide

Example: Let’s say that we start at a position of 10

m, we end up at 50 m, out initial velocity is -20 m/s, and our acceleration is 2 m/s2. At what time do we arrive? Let’s plug everything in!

x = xo + v(t) + or 50 = 10 + -20(t) + That simplifies to: 40 = -20(t) + 1(t2) How can we solve for t and t2? We must learn a new formula!

The Quadratic Equation!!! This is one of the most fundamental

equations you will learn in math. It allows us to solve these kinds of problems every time! Here it is:

If at2 + bt + c = 0, then Does anyone know what means?

? What does mean? It means that there are two possible

answers! One in which you add to get the answer, and one in which you subtract!

So if I ask what is 5 3, you would do both 5 + 3 and 5 – 3.

The answers would be 8 and 2!

Ok, let’s try that example problem!

We know that If at2 + bt + c = 0, then , and we simplified to 40 = -20(t) + , so now we need to rearrange what we have to fit the first formula. It is:

1(t2) – 20(t) – 40 = 0 Now all we need to do is plug our

numbers into the second equation!

Example Cont. We know that If at2 + bt + c = 0, then , and 1(t2) – 20(t) – 40 = 0 Then = = = = 21.83 s and – = –1.83 s

Answers! Try it if you like, but –1.83 and 21.83 do

satisfy the original equation (40 = -20(t) + )

Do both answers make sense? Can we have a negative time? NO!!! That leaves us with 21.83 seconds as

our final answer!

Your turn! Clicker Question – Let’s say that we start at a position of 5

m, we end up at 10 m, out initial velocity is 2 m/s, and our acceleration is 2 m/s2. At what time do we arrive?

A. 2.89 B. -3.44 C. 3.44 D. 1.44 E. 1

Are there any answers we can get rid of automatically?

A. 2.89 B. -3.44 C. 3.44 D. 1.44 E. 1 YES! We can’t have a negative time, so

we automatically know that B is wrong! Let’s do the problem and see which one

is correct!

Answer Cont. We know that If at2 + bt + c = 0, then , and 1(t2) + 2(t) – 5 = 0 Then = = = = 1.44 s and – = –3.44 s

Correct answer: 1.44 A. 2.89 B. -3.44 C. 3.44 D. 1.44 E. 1 D is our correct answer!