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What I want to do with this video

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is think about what happens to some type of projectile,

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maybe a ball or rock, if I were to throw it

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straight up into the air.

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To do that I want to plot distance relative to time.

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There are a few things I am going to tell you about

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my throwing the rock into the air.

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The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s)

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I picked this initial velocity because

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it will make the math a little bit easier.

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We also know the acceleration near the surface of the earth.

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We know the force of gravity

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near the surface of the earth is the mass of the object times the acceleration.

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(let me write this down)

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The force of gravity is going to be the mass of the object times little g.

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little g is gravity near the surface of the earth

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g is 9.8 meters per second squared (9.8m/s^2)

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Now if you want the acceleration on earth

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you just take the force divided by the mass

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Because we have the general equation

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Force equals mass times acceleration (F=ma)

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If you want acceleration divide both sides by mass

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so you get force over mass

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So, lets just divide this by mass

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If you divide both sides by mass,

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on the left hand side you will get acceleration

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and on the right hand side you will get the quantity little g.

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The whole reason why I did this is

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when we look at the g it really comes from

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the universal law of gravitation.

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You can really view g as

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measuring the gravitational field strength near the surface of the earth.

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Then that helps us figure out the force

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when you multiply mass times g.

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Then you use F=ma, the second law,

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to come up with g again

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which is actually the acceleration.

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This is accelerating you towards the center of the earth.

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The other thing I want to make clear:

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when you talk about the Force of gravity

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generally the force of gravity is equal to big G

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Big G (which is different than little g) times

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the product of the masses of the two things

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over the square of the distance between the two things.

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You might be saying "Wait, clearly the force of gravity is dependent on the distance.

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So if I were to throw something up into the air,

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won't the distance change."

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And you would be right!

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That is technically right, but

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the reality is that when you

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throw something up into the air

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that change in distance is so small

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relative to the distance between the object and the center of the earth

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that to make the math simple,

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When we are at or near the surface of the earth (including in our atmosphere)

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we can assume that it is constant.

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Remember that little g over there is

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all of these terms combined.

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If we assume that mass one (m1) is

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the mass of the earth, and

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r is the radius of the earth (the distance from the center of the earth)

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So you would be correct in thinking that it changes a little bit.

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The force of gravity changes a little bit, but

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for the sake of throwing things up into our atmosphere

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we can assume that it is constant.

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And if we were to calculate it

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it is 9.8 meters per second squared

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and I have rounded here to the nearest tenth.

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I want to be clear these are vector quantities.

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When we start throwing things up into the air

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the convention is

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if something is moving up it is given a positive value,

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and if it is moving down we give it a negative value.

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Well, for an object that is in free fall

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gravity would be accelerating it downwards, or

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the force of gravity is downwards.

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So, little g over here,

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if you want to give it its direction,

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is negative. Little g is -9.8m/s2.

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So, we have the acceleration due to gravity.

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The acceleration due to gravity (ag) is

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negative 9.8 meters per second squared (9.8m/s^2).

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Now I want to plot distance relative to time.

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Let's think about how we can set up a formula,

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derive a formula that, if we input time as a variable,

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we can get distance.

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We can assume these values right over here.

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Well actually I want to plot displacement over time because that will be more interesting.

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We know that displacement is

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the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)).

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Right now we have

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something in terms of time, distance, and average velocity

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but not in terms of initial velocity and acceleration.

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We know that average velocity is the same thing as

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initial velocity (vi) plus final velocity (vf) over 2. (Vavg=(vi+vf)/2)

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If we assume constant acceleration.

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We can only calculate Vavg this way assuming constant acceleration.

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Once again when were are dealing

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with objects not too far from the center of the earth

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we can make that assumption.

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Assuming that we have a constant acceleration

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Once again we don't have what our final velocity is.

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So, we need to think about this a little more.

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We can express our final velocity in terms of our initial velocity and time.

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Just dealing with this part, the average velocity.

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So we can rewrite this expression

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as the initial velocity plus something over 2.

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and what is final velocity?

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Well the final velocity is going to be

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your initial velocity plus your acceleration times change in time.

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If you are starting at 10m/s

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and you are accelerated at 1m/s^2

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then after 1 second you will be going 1 second faster than that. (11m/s)

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So this right here is your final velocity.

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Let me make sure that these are all vector quantities...(draws vector arrows)

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All of these are vector quantities.

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Hopefully it is ingrained in you that these are all vector quantities, direction matters.

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And let's see how we can simplify this

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Well these two terms

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(remember we are just dealing with the average velocity here)

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These two terms if you combine them become 2 times initial velocity (2vi).

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two times my initial velocity

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and then divided by this 2

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plus all of this business divided by this 2.

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which is my acceleration times my change in time divided by 2.

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All of this was another way to write average velocity.

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the whole reason why I did this is because we don't have final velocity

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but we have acceleration

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and we are going to use change in time as our independent variable.

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We still have to multiply this by this green change in time here.

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multiply all of this times the green change in time.

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All of this is what displacement is going to be.

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This is displacement, and lets see...

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we can multiply the change in time times all this

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actually these 2s cancel out

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and we get (continued over here)

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We get: displacement is equal to

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initial velocity times

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change in time

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Some physics classes or textbooks put time there but it is really change in time.

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change in time is a little more accurate

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plus 1/2 (which is the same as dividing by 2)

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plus one half times the acceleration

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times the acceleration

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times (we have a delta t times delta t)

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change in time times change in time

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the triangle is delta and it just means "change in"

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so change in time times change in time

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is just change in times squared.

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In some classes you will see this written as

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d is equal to vi times t plus 1/2 a t squared

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this is the same exact thing

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they are just using d for displacement

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and t in place of delta t.

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The one thing I want you to realize with this video

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is that this is a very straight forward thing to derive.

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Maybe if you were under time pressure you would want to be able to whip this out,

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but the important thing, so you remember how to do this when you are 30 or 40 or 50

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or when you are an engineer and you are trying

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to send a rocket into space and you don't have a physics book to look it up,

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is that it comes from the simple displacement is equal to average velocity times change in time

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and we assume constant acceleration,

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and you can just derive the rest of this.

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I am going to leave you there in this video.

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Let me erase this part right over here.

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We are going to leave it right over here.

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In the next video we are going to use this

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formula we just derived.

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We are going to use this to actually

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plot the displacement vs time

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because that is interesting and we are going to be thinking about

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what happens to the velocity and the acceleration

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as we move further and further in time.