WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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Today I want to talk about defining and graphing motion.
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Our objectives are going to be to understand the difference between position, distance, and displacement.
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Our second objective is to understand the difference between speed and velocity.
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Our third objective is to solve problems involving average speed and velocity and calculating distance displacement, speed velocity, and acceleration.
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Our fourth objective will be constructing and interpreting graphs and diagrams of position, velocity, and acceleration versus time.
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And finally, we will be determining and interpreting slopes and areas of motion graphs in order to help us understand what these quantities are.
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So, with that, let us dive right in. Let us first define position.
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An object’s position in a one-dimension -- you can assign it to a variable on a number scale.
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Very simply, if we make this an x-axis, for example, we can put an object’s position anywhere on that x-axis and it describes where that is at a specific point in time.
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Ha! There it is. You can assign whatever you want to be the 0-point as well as the positive and negative directions.
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Now the symbol for positioning in one-dimension is x.
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If we look at distance when position changes, an object has traveled some amount of distance.
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The more position changes, the more distance is traveled.
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Distance is a scalar. It has a magnitude only -- no direction.
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And its standard measurement is in meters (m). Distance is a scalar -- no direction.
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So let us take a look at an example here.
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We have a deer that walks 1300 m East to a creek for a drink.
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The deer then walks 500 m West to the berry patch for dinner.
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Then it runs 300 m West when startled by a loud, fierce raccoon.
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What total distance did the deer travel? Well, in order to do this, let us just take a look.
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The deer traveled 1300 m East, then 500 m West, and then 300 m West.
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So the total distance traveled, 1300 + 500 + 300, should be right around 2100 m.
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Very straightforward. No direction required.
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Displacement, on the other hand, is a vector which describes the straight line distance from where you start to where you end.
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It includes a direction. It is a vector.
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Displacement, final x position minus initial x position or we also call that Δ (Δ) x is also measured in meters.
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So let us take a look at an example now with displacement.
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Same basic story, slightly different take on it.
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Now our deer walks 1300 m East to the creek, then walks 500 m West to the berry patch, and then runs 300 m West when startled by a loud raccoon.
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What is the deer's displacement? Well to find the displacement, we go from a starting point to our final point.
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He went 1300 m East and then 500 West and 300 West.
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Well, the total from the starting point to its final point, must be 500 m East.
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So the deer's displacement, 500 m East.
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Displacement is a vector, therefore, it must have a direction as well.
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Now average speed tells you the distance traveled divided by the total time it took to travel that.
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And we give it the symbol v with a line over it. The line over it meaning average.
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So that is the distance traveled divided by the time.
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Speed is a scalar -- S-S and its measured in meters per second (m/s).
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So let us take our same example and let us look at average speed.
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The deer walked 1300 m East, then 500 m West, and then ran 300 m West.
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It did that entire trip in 600 seconds (s).
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If that is the case, to find the average speed for the entire trip, v-average, average speed equals total distance traveled divided by the time it took.
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We already found the total distance traveled is 2100 m and the time it took, 600 s.
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Just a little bit of math and I find out that the average speed was 3.5 m/s.
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Moving on, average velocity is the displacement divided by the time interval.
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Since it is a displacement and displacement is a vector, average velocity is also a vector.
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It is also measured in m/s, but it includes a direction.
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Average velocity, again same symbol, v with a line over it, so we have got to be careful.
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It is x - x initial over the time it took or Δx/T.
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Let us look at that in the context of our same example as well.
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Deer walked 1300 m East to the creek, 500 West for dinner, and then ran 300 m West.
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What is the average velocity, the vector, if the entire trip took 600 s?
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Once again, we are going to start off with the same sort of math.
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The average is going to be equal to the displacement divided by the time.
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But now our displacement is 500 m East. The time it took was 600 s.
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500/600 is going to be 0.833 m/s and since it is a velocity -- it is a vector -- it must have a direction to go along with it.
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Let us see if we cannot put all of that together.
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Here is our problem with our dear friend, Chuck, the hungry squirrel.
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Chuck, the hungry squirrel, travels 4 m East and 3 m North in search of an acorn.
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The entire trip takes him 20 s. Find the distance he travels, his displacement, his average speed, and his average velocity.
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Well distance traveled, if he went 4 m East and 3 m North, how far did he go?
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He went 7 m -- distance --7 m.
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What is his displacement? Well to do that we need to figure out how far he went from his starting point to his ending point.
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He started off down here. He went 4 m East, then he went 3 m North.
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His displacement then, is going to be the vector from the starting point of the first to the ending point of our last.
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There is our displacement. That is a 3/4 5 triangle or you could find the magnitude of that by the Pythagorean Theorem.
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You should still come up with 5 m. So his displacement, Δx must be 5 m and direction -- he went Northeast.
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How about his average speed? Again, average speed is going to be distance divided by time.
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His distance we just found was 7 m. It took 20 s, so that is going to be 0.35 m/s for his average speed.
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Average velocity, on the other hand, is going to be his displacement divided by time.
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Displacement was 5 m to the Northeast in 20 s or 0.25 m/s Northeast.
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It has a direction, it is a velocity.
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So note very carefully here, average speed and average velocity do not have to have the exact same numerical value, even when you are talking about very similar paths or the same paths.
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All right -- acceleration. Acceleration is the rate at which velocity changes.
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If everybody just kept going a constant velocity, the same speed, it would be a pretty boring world.
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So when velocity changes, we call it an acceleration. It is the change in velocity divided by the time.
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And Δv or Δanything is always the final value minus the initial value.
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Acceleration, too, is a vector. It has a direction and its units are meters per second per second.
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What that means is if your speed changes -- if your acceleration is 5 m/s/s, your speed increases by 5 m/s every second.
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I should say your velocity changes 5 m/s, every second.
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We also call that a meter per second squared (m/s ²).
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All right. Let us take a look at an example with acceleration.
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Monte, the monkey, accelerates from rest, so his initial velocity is 0, to a velocity of 9 m/s.
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V = 9 m/s in a time span of 3 s. Find Monte's acceleration.
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Now acceleration is change in velocity divided by time or final velocity minus initial velocity divided by time.
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That is going to be 9 m/s - 0/3 s or 3 m/s/s, which we would typically write as 3 m/s ².
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Very straightforward. A simple problem in acceleration, but hopefully it starts giving you the idea.
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Now, we have to talk a little bit about average values versus instantaneous values. They are not the same thing.
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Average values take into account an entire time interval.
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Instantaneous time values tell you the rate of change of a quantity at a specific instant in time at that specific point.
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They are not always the same. So let us take a look at a problem with average velocity.
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A motorcyclist travels 30 km in 20 minutes. 30 km in 20 minutes -- there is part of our trip -- at a constant velocity.
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He takes a 10 minute break, then travels 30 km in 30 minutes at a constant velocity.
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Find the cyclist's minimum instantaneous velocity, maximum instantaneous velocity, and average velocity.
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It is pretty easy to see the minimum instantaneous velocity is going to occur here where they are taking a break.
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Position is not changing -- velocity is 0. So the minimum instantaneous velocity is 0.
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Where are we going to have the maximum instantaneous velocity?
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Well, that is going to be when we travel the biggest distance in the smallest amount of time, or up here we are traveling 30 km in 20 minutes.
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So the max is going to be 30 km/20 minutes or 1.5 km/min.
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Now to find the average velocity though, we have to take into account the entire trip.
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Average velocity is going to be the total displacement, 60 km, divided by the total time.
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20 + 10 + 30 is 60 minutes, which will be 1 kilometer per minute (km/m).
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And note here that the average velocity is between the minimum and maximum -- always going to be either between or equal to the minimum or maximum.
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It cannot be outside that and be a real average velocity. A great check on your problem solving.
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Let us take a look at particle diagrams.
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It is kind of similar to the effect you might see if you had an old leaky car that has an oil leak and every second, every specific instance in time, it leaks one drop of oil.
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It is a consistent drop every specified unit of time.
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So if we were traveling down the road to the right in our car at a constant speed, we would see oil drops hit the ground evenly spaced.
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Because they are evenly spaced, you know the car is moving at a constant velocity.
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So just by looking at the oil drops you would be able to go take a look and see exactly what is happening.
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Now when you looked at these, there are a couple of questions that might remain in your mind.
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You are pretty certain the car is traveling at constant velocity, but can you tell if it is traveling to the right or to the left just from the oil drops.
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Now assuming that you were not looking on which side of the road they were on, you really cannot tell.
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But you do know that regardless, the acceleration is 0.
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It is moving at constant velocity because these are all evenly spaced -- same spacing between each of the drops.
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On the other hand, what if our car was accelerating to the right?
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Our particle diagram is now non-uniformed. We have smaller spacings over here and bigger spacings over here.
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Drops are getting further apart, velocity is changing, the car is accelerating.
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Can you think of a case in which the car could have a negative velocity and a negative acceleration, yet speed up?
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Think about it for a second.
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Here is our car -- put some wheels on it and make it a nice, pretty little car. There it is.
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If it has a velocity to the left and we have called to the left negative, that would be a negative velocity.
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If it is accelerating in that direction, it is going faster and faster and the velocity to the left is getting bigger and bigger in magnitude, but that is a negative acceleration.
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Negative acceleration does not mean slowing down. It is not decelerating.
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Negative acceleration just means that you are accelerating in whatever direction you have called negative.
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If you have velocity and acceleration in the same direction, the car will be speeding up.
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Its speed will be increasing.
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Let us take a look at some motion graphs.
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One of the most popular types is a position time graph. It shows an objects position as a function of time.
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So let us assume that we have some cute little dog that wanders away from our house at a constant 1 m/s.
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So the dog does that -- starts at time 0 -- wanders away from the house for a little bit, so the position is getting further and further away from this origin -- the house -- until here the dog decides it has had enough and takes a five second rest.
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It bops down in the grass in the backyard for 5 s. Its position does not change.
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Then the dog returns to the house at 2 m/s.
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So it is going back the opposite direction and we end up with a little bit steeper slope because it is coming back to the house faster.
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There is a basic position time graph. We can learn an awful lot from this graph though.
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The slope of this graph tells you the dogs velocity at any given point in time.
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For the first part of the trip, if we take the slope of the graph -- right here -- the slope there is going to be rise over run.
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We rise 1, 2, 3, 4, 5 meters in a time of 1, 2, 3, 4, 5 seconds or 1 m/s. That is the dog's velocity. Remember?
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So the slope of a position time graph, gives you the velocity.
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Let us take a look here when the dog's taking a rest.
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Slope of a flat line is 0. The dog's velocity was 0.
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Let us take a look here coming back to the house over here on the right.
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The slope there again, rise over run. Our rise now is 1, 2, 3, 4, 5, but it is in the opposite direction.
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It is going down, so our rise is -5 meters and the time it took goes from 10, 11, 12, 12 1/2 -- 2.5 seconds.
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So I come up with a slope of -2 m/s.
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The velocity of the dog is -2 m/s -- where that negative sign -- all that is telling you is that the negative is indicating the direction.
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So position time graphs can be a very useful tool for describing the motion of an object.
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We could also make a velocity time graph. It shows the velocity of an object as a function of time.
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It is related to the position time graph by the slope.
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So here is the position time graph we were just talking about for the dog wandering away from the house and back.
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Down below we have the velocity time graph for basically the same information.
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Initially, the dog's velocity, the slope here was 1 m/s, so our velocity down here for that same time interval of 5 s is 1 m/s.
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Then the dog took a rest for a couple of seconds -- velocity was 0 -- slope was 0.
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Here we have a value on our velocity time graph of 0 for that same time interval.
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Now to end the story, the dog came back to the house -- our slope was -2 m/s.
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Our velocity down here is -2 m/s for the same time interval, so these graphs are very closely related.
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We can look at the area under the velocity time graph to tell us the change in the displacement of the dog even.
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Here is how that works.
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As we look at the area under the graph of our velocity time graph -- if we take this area, the velocity under the graph, the area between the 0 line and where our graph is, we get that rectangle.
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The area of that rectangle is length times width.
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Our length is 5 s. Our width is 1 m/s, which is 5 m.
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By the time we get to 5, our area is 5 m, but a look at our position time graph, right at that point, the position is 5 m.
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If we keep going through our graph and say "Hey over here at 8 s, what total area do we have?"
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All the area to the left of that 8 s is still 5 m, so over here at 8 s, our position is still 5 m.
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And if we keep going -- if we wanted to say what is the dog's position over here at 12 1/2 s -- well -- now we have another area to take into account.
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We also have this rectangle and since it is below the line -- although in math they may tell you there is not officially a negative area, there is a meaning to negative areas on the graphs in physics -- that area, which is going to be length times width -- we have 2 1/2 s width and we have -2 m/s.
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So 2 1/2 x -2 is going to be -5 m.
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So all of the area added up to the left of this red line -- we have +5 and -5 gets us 0.
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But look, down here at exactly that point in time. 12 1/2 s, just like we have here, our position is back to 0.
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Now the slope of the VT graph, the velocity time graph, can also give you a lot of information.
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It can give you the acceleration. So with a problem like this, we can also look and say, "What is the acceleration of the car at T = 5 s?
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Well at T = 5 s -- what we have is a slope of 0.
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So the acceleration at that point -- 0 m/s ².
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How about the total distance traveled by the car during the 6 s interval?
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Well, to get the total distance traveled, we need to take the area under all of this.
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How do you take the area of that? A couple of ways to do it, but probably the easiest in my mind is I would break this up into a couple of shapes.
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Over here we have a triangle and on the right hand side we have a rectangle.
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Let us add up their areas to find the total distance traveled.
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Total distance traveled is going to be the area of our red triangle, 1/2 base times height, plus the area of our blue rectangle.
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One-half our base is 4 s. Our height is 10 m/s and our rectangle -- our length is from 4-6 -- 2 s, and a height of 10 m/s.
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So 1/2 x 4 x 10 -- that is going to be 20 m + 2 x 10 -- 20 m -- for my grand answer of 40 m traveled.
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All right -- how we can use slopes and areas.
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Now, acceleration time graphs -- you get by taking the slope of the velocity time graph.
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So, if you take the area under the acceleration time graph, you get the objects change in velocity.
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We have a pattern here. We started with position time graphs.
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If we took the slope, we got velocity.
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If we have a velocity time graph and we take the slope, we get acceleration -- or the other direction -- if we take the area under the acceleration time graph, we get the change in velocity.
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From the velocity time graph, if we take the area, we get the change in position.
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You can keep going with this pattern forward and backwards to find whatever information you need to based on these motion graphs.
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Let us take a look at another example with a position times, sometimes distance time graph.
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Which graph here best represents the motion of a block accelerating uniformly down an inclined plane?
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Well let us think about what is going to happen if we have some inclined plane or a ramp and we have a block that is accelerating down the ramp.
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Initially it is going to be at position 0 and over time it is going to get a larger and larger distance traveled.
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So, right away, we can eliminate number one on our choices.
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Now, as it goes faster and faster, it is going to cover more and more distance.
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We would also think of that as seeing that. . . .
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If we looked over here at a velocity time graph, we could make a velocity time graph and say "You know, it probably starts at some 0 velocity and goes faster and faster and faster."
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Well if that is the case, we also need to look at something where the area is getting progressively bigger, therefore the distance traveled must be getting progressively greater for the same time interval.
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The correct answer must be that one.
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And we could also look at it from the opposite direction. Right here, the slope is 0, so we have a velocity of 0.
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Here we have a bigger slope. We have a bigger velocity.
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Here we have a very big slope. We have a very big velocity as time increases.
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Therefore, that would be the graph that best represents the motion of a block accelerating uniformly down an inclined plane.
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Let us take a look at a little bit of a challenging one. See if we cannot push a little bit.
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Bobbie dribbles the basketball on the ground.
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Draw the position time graph, the velocity time graph, and the acceleration time graph for the basketball as it travels down from Bobbie's hand -- bounces back up to her hand.
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We will assume the floor is position Y = 0.
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So let us start by making a graph of position Y versus time.
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Let us make one for velocity versus time and let us make one for acceleration versus time.
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All right. Initially the ball is going to start off in Bobbie's hand.
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So position-wise, it is going to start up here at some positive value.
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We also know after a little while, it is going to hit the ground -- position 0 -- and as it comes back up to Bobbie's hand, it is going to come back to where it started.
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So we have those points for position.
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Now for velocity, let us assume that Bobbie drops it as opposed to really pushing it, just to make it a little simpler.
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That means its initial velocity is going to have to be 0.
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As it is traveling down towards the ground, it is going to go faster and faster.
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Then the moment it hits the ground, its velocity switches its direction, but keeps roughly the same magnitude.
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So we are going to have to have this jump back up here and as it comes back up to Bobbie's hand, it gets slower and slower and slower until its velocity is 0 right at her hand level.
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So we are going to have to have something like that and we will fill in some of the other points in a minute.
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As far as acceleration goes, the entire time the objects in the air and nothing's touching it, it is acceleration is the acceleration due to gravity here on Earth.
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It is constant. It does not change.
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It is -9.8 m/s ², so it is going to be constant -- then that ball is going to come in contact with the floor.
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Its acceleration is going to change very quickly and then it is going to be in -- we call free-fall -- again.
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It is going to be in the air with no other forces on it.
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It is going to have a constant 9.8 m/s ² again.
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All right. How can we start to fill these in?
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Well, when the ball hits the ground, its acceleration is going to be positive for a second.
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It has to be in that direction to change its velocity.
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So we are going to have to have a spike in our acceleration time graph.
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For our velocity graph, it is going to start at 0 and it is going to go faster and faster and faster.
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Then, when it hits the ground it is going to have a spike.
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It is going to have a very high velocity as it sways back up, slowing down, slowing down, slowing down -- stopping.
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And finally, as we look at the position of the basketball -- if we take a look, we have to have some sort of path that allows the ball to do that to come back up to Bobbie's hand.
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So that is a pretty in-depth example and much more complicated example of how you can put position, velocity, and acceleration all together to make one complete story for what is happening to an object.
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All right, let us do another one.
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Draw the velocity time graph for a ball tossed upward which returns to the point from which it was tossed.
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Well, I am going to start off by making my axis again. This is going to be a velocity time graph.
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If we toss something upwards -- like throw it up -- the moment it leaves my hand, it has its biggest velocity. Right?
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It's positive -- slowing down, slowing down, slowing down, slowing down -- stops for a split second, switches directions, speeds up, speeds up, speeds up, speeds up, speeds up, but in the negative direction.
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So if it starts off with its biggest velocity -- it could be there -- a little bit later at its highest point, for a split second it stops, then it goes faster, and faster, and faster in the opposite direction.
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So the velocity time graph for that situation would look something like that.
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All right, let us take a look at one last example.
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How can we get displacement from a velocity time graph?
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The graph below shows the velocity of an object travelling in a straight line is a function of time.
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Determine the magnitude of the total displacement of the object at the end of the first 6 s.
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So we have a velocity time graph -- we want displacement. Right away you should be thinking area.
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Velocity time graph wants displacement -- you need to take the area.
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So at 6 s, we will draw our line there.
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We need the area of everything under the graph to the left of that.
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Again, a couple of ways you can do this -- but the easiest way I see it off the top of my head is to break this up into a triangle and a rectangle.
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The area under that should give us the total displacement.
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So we have the area of the triangle, 1/2 base times height or 1/2 times our base 2 s times our height of 10 m/s is going to be 1/2 x 2 x 10 -- 10 and seconds versus seconds in the denominator -- meters.
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And the area of our rectangle, length times width, or from 2 to 6 s is 4 s times its height -- 10 m/s -- seconds over seconds cancel out -- 40 m -- so the total displacement then, I just add those two up, 40 + 10 -- 50 m.
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Hopefully, that gets you started with some of these quantities that describe motion and motion graphs, particle diagrams, position time diagrams, velocity time diagrams, acceleration time diagrams -- gets you started, gets you going.
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I definitely recommend some more practice on your own.
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