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 ourposition over time?

It simply means that we move from onepoint to another in a certain amount oftime.

Remember, we measure velocity inmeters/second, or m/s.

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

Maybe if we just use the same idea, wecan get the right units for this thing calledacceleration.

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

If Velocity = Meters Seconds , then dividing by seconds will look like:
Acceleration = Velocity Seconds or:
Acceleration = Meters Seconds°Seconds
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 = Velocity Seconds , not Meters Seconds°Seconds
Although, we do abbreviate the units as: m/s2 .

We thought of velocity as the time it tookto get from one position to another, sowe will think of acceleration as:

The time it takes to get from one velocityto another!

A positive acceleration means that weare speeding up in the positive direction,while a negative acceleration means weare speeding up in the negativedirection.

Let’s think about this a little deeper –

We know that a constant velocitychanges something's position, but howdoes acceleration change things…Doesit change velocity and position?

YES! It changes both!

In a 2 dimensional world, we have 4possible ways to think about velocity andacceleration: 1 where both are positive,1 where both are negative, and 2 wherethey are opposites!

Acceleration gets a blue arrow,while velocity get the red arrow

Lets talk about the examples where theyare both the same, either both positiveor negative

In this example, we are moving forward(positive velocity,) but every second thatwe are moving, we are moving forwardfaster! (positive acceleration)

Negative velocity & Negative acceleration:

Now, we are moving in the negativedirection (negative velocity,) but everysecond that we are moving, we aremoving in the negative direction faster!(negative acceleration)

Now, we are moving in the negativedirection (negative velocity,) but everysecond that we are moving, we arespeeding up in the positive direction!(positive acceleration)

Whoa, what does that mean???

If we are speeding up in the oppositedirection we are moving, then we slowdown (until net velocity is zero,) then we willactually move in the direction of theacceleration – positive direction in this case

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

Even though our acceleration and velocityare in opposite directions, if we arespeeding up in the opposite direction weare moving, then we slow down (until netvelocity is zero,) then we will actually movein the direction of the acceleration – thenegative direction in this case!

Yes it was, but we still need toaccomplish two goals:

What equation can we use to find thefinal position of a moving object with aconstant acceleration?

We already know an equation formoving under constant velocity:

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

Don’t forget how we figured it out!!!

We know that velocity is measured inm/s, so if we want to convert to position(just meters,) all we need to do is multiplyby a time (whichever time we want touse) to turn it into a position!

Maybe we should try this again foracceleration?

YES!!!

We already have our first equation:

x = xo + v(t)

Now we just need to add in a part thatdeals with acceleration!

How can we turn acceleration into aposition?

Well, we already know how to turn avelocity into a position, so we just needto find a way to turn acceleration into avelocity!

Well, it shouldn’t be that hard, we already know that:
Acceleration = Velocity Seconds , 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!

No, unfortunately it is not quite that easy.

It actually took a long time to figure outwhy that didn’t work.

The reason comes from one of the mostcomplicated 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!

In reality, we were actually very close to being right the first time!
It is: x = xo + v(t) + a(t2) 2
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.

Yes, after we finish with this powerpoint,you will do some simulations to help youcement your understanding.

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

This is harder than you might think!

Let’s do an example problem on thenext slide

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) + a(t2) 2 or 50 = 10 + -20(t) + 2(t2) 2
That simplifies to: 40 = -20(t) + 1(t2)
How can we solve for t and t2?
We must learn a new formula!

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 t = −b ± b 2 − 4(a)(c) 2(a)
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!

We know that If at2 + bt + c = 0, then t = −b ± b 2 − 4(a)(c) 2(a) , and we simplified to 40 = -20(t) + 2(t2) 2 , 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!

We know that If at2 + bt + c = 0, then
t = −b ± b 2 − 4(a)(c) 2(a) , and 1(t2) – 20(t) – 40 = 0
Then t = 20 ± 20 2 − 4(1)(−40) 2(1) =
t = 20 ± 400 + 160 2 = 20 ± 23.66 2 =
43.66 2 = 21.83 s and – 3.66 2 = –1.83 s

Try it if you like, but –1.83 and 21.83 do satisfy the original equation (40 = -20(t) + 2(t2) 2 )
Do both answers make sense?
Can we have a negative time?
NO!!!
That leaves us with 21.83 seconds as our final answer!

Let’s say that we start at a position of 5m, we end up at 10 m, out initial velocityis 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