Okay, acceleration and gravity. Here's where it gets interesting. In the previous lesson, we only talked about constant speed cases, and of course, in the real world, very few things are moving at a constant speed. When speed changes, that's acceleration. In every day language, acceleration simply means speeding up.

But in Physics, acceleration means any change in speed, getting fast or getting slower. Those are two different kinds of acceleration. Sometimes we call the getting slower, deceleration. But that's just a kind of acceleration. We will consider only uniform acceleration.

That means that the acceleration itself is not changing. So, if the speed is changing, but the acceleration itself is staying constant. Now, what's tricky is that if acceleration starts changing, well then, all the equations have to involve Calculus, then we're doing Physics with Calculus. And Physics with Calculus is a little beyond what the MCAD is gonna ask of you. And so, on the MCAD, you will see constant acceleration.

You're not gonna see a whole lot of changing acceleration. The fundamental equation for constant acceleration is just acceleration equals change in speed over time. So V final minus V initial over t. And this can be rewritten in this form, so that is just a rewritten form of the definition of acceleration.

And this is one of the fundamental kinematic equations. So, store that in your memory banks, we'll go back to that, that's kinematic equation number one right there. V final equals at plus V initial. If distance is in meters and time is in seconds, then speed would be in meters per second.

So, meters per second is the fundamental metric unit of speed and incidentally I'll say, just for approximation purposes. If you take a speed in meter per second and double the number that's approximately the speed in miles per hour. So, something's going 30 meters per second, that's approximately 60 miles an hour, that's just a good approximation to get a ballpark idea about speed.

What are the unites of acceleration? Well, an acceleration is speed divided by time, so that's meters per second divided by second, and we often write that as meters per seconds squared. And that's something that can be confusing to people, what is does that mean? Meters per seconds squared. Well, example an acceleration of 3 meters per second squared means that for every second the speed is increasing by an additional 3 meters per second.

So, during the first second, let's say it starts out at rest. After one second it the object is going 3 meters a second. After two seconds it's going 6 meters a second. Then 9. Then 12. Then 15.

Then 18 meters per second. So that speed is climbing and that's exactly what the acceleration of 3 meters per second means. One way to think about this stuff is graphically. Uniform acceleration, that is acceleration, constant acceleration. Acceleration, a constant value, appears as a straight line on a graph of speed versus time.

And so, you can see what's going on here, the slope is the acceleration. And so what's happening is with this line, of course, the speed is going up and up and up, but the slope is remaining constant. That is the constant acceleration. Here, the slope is going down, so the speed is going down, down, down but again the slope is constant.

That's the constant acceleration. Notice this is on a speed versus time graph. Now, instead if we have a distance versus time graph, an x versus t graph, then acceleration appears as a parabola. On this graph, the slope is the speed. So notice what's happening.

Here, we're going from low slopes to steeper and steeper slopes. The graph is getting steeper and steeper. So that means the speed is increasing. Here what's happened is the slope starts out steep and then it gets less and less steep. So, here we have deceleration.

So, acceleration is an upward curving parabola, deceleration is a downward curving parabola. At least when it's moving up like this. Now constant speed appears as a straight line on as an x-vs-t graph. And appears as a horizontal line as a v-vs-t graph. So notice, it's very important.

You can't just relay on the shape to tell you what the motion is. You have to pay attention to whether it's a speed versus time graph or a distance versus time graph. So these two graphs here, these are shown exactly the same motion, constant velocity motion, but on two different graphs. On the v versus t graph, because the speed is constant, it's just a horizontal line.

As time goes on, it's staying at that one speed. Whereas on the x versus t graph, obviously position is changing, but the slope, the speed, is remaining constant. Again, you can't just rely on shape, you have to pay attention to whether it's a v versus t or an x versus t graph. So, here's a practice problem.

Pause the video, and then we'll talk about this. Okay, let positive x indicate the direction a car is facing. Which of the following distance vs time graphs would represent a car in reverse accelerating backwards? So this is the driver who puts the car in reverse and then steps on the gas, so the car accelerates backwards.

Well, let's see. What's going on in A? So notice, first of all, any of the graphs where it's going from down to up, the car is moving forward. So A, what's happening is it's moving forward, and we also have slope increasing.

So this is accelerating forward. A would be foot on the gas, car in drive. Notice that C is also going up, but here we have deceleration. We go from high slope to a lower slope. So this would be car on drive, but foot on the brake. Decelerating forward.

So, it's gonna be either B or D cuz both of these are in drive, they're not in reverse. We look at B, what's happening is, we have a big negative slope, a negative slope that is very steep, and it's getting less and less steep in going to zero. So in other words this is deceleration backwards, this is car in reverse, put on the brake.

Where as in D, we start out with a low slope and then the slope gets more and more negative. So, it's going more and more in the backwards direction. So this is the car in reverse, foot on the gas and this is the one we want, car in reverse accelerating backwards, answer choice D. Now you may remember from the last video, if we have two different stretches of constant velocities at two different speeds, we can't simply average the two speeds to get the average velocity.

And so, if you remember the practice problem from last time, this is actually a graph of that practice problem. For six hours, going at 20 miles an hour, and then for two hours going at 60 miles an hour. And of course, what we have to do is figure out the individual distances, individual times, in order to figure out the average velocity.

We did that in the last lesson. By contrast, when a speed uniformly increases over the interval, when you have constant acceleration. We can numerically average the individual final velocities. So here we have a v versus t graph of something that is accelerating. It starts at 30 meters per second, and then over the course of eight seconds, it actually accelerates to 70 meters a second.

And so this is the basic setup, from this notice we could figure out the acceleration. That would be change in velocity divided by time. Five meters per second squared. That's a pretty good acceleration right there. If you could get a car that accelerates at five meters per second squared that's an awfully decent acceleration.

Notice also, we can average the beginning and the ending velocities to get the average velocity. The average velocity of this entire trip is going to be 50 meters a second. And so this is a very different case. If we have two plateaus as we saw in the previous, we can't just average those velocities.

But if we have a constant smooth straight line, the average height of this straight line is the average of the starting point and the ending point. And that's why we can average the starting point and the ending point to get the average velocity. So far, we have two important kinematic equations. So the first one I re-wrote it a bit, v final minus v initial equals AT.

The second one, average velocity equation, total distance divided by time equals v initial plus v final divided by two. Now notice we can multiply these to cancel t. So I'm gonna multiply them like this. Multiply the AT times x total divided by t and multiply that difference, v final minus v initial by that fraction.

What happens, of course, the t's cancel and on the left we're gonna multiply both sides by 2, just to move that 2 in the denominator to the other side. And we get the difference of two squares, that algebraic pattern a minus b times a plus b equals a squared minus b squared. And that gives us this equation. That's a really important equation.

Sometimes this is called the time independent equation because this equation has all the other kinematic factors but it doesn't have time. So that is the third important kinematic equation, again, called the time independent equation. Start again with those two equations. And what we're gonna do this time is plug the left equation into the right equation.

And so, we'll plug in like this, we get 2 of the v initials. Then what we're gonna do is separate out the fraction, so we get 1/2 at + v initial, and then just multiple both sides by t. So we get X total equals one half AT squared plus ViT. That's the fourth big kinematic equation.

And so these four equations are four equations you need to know. And so here they are again. And notice what's important about each equation is not only noticing what values are in each equation, but also being aware of what is not in each equation. Each one of these is missing a different thing. And so, there's a problem where we don't care about acceleration, we might use equation 2.

If we don't know and don't care about the final velocity, we might use equation 4. And so it's very important to be aware of not only what each equation has, but also what it doesn't have. Part of the basic strategy in any kinematic problem is to discern which equation to use from what you're given, what you need to find, and also what you are not given and what you don't care about.

All of those together go into the choice of the right kinematic equation. Here's a practice problem. Pause the video, and then we'll talk about this. Okay. A car, initially traveling at 20 meters per second, decelerates to rest at negative four meters per second squared.

How much distance is covered during this deceleration? So a few things. Notice we know initial velocity, we know an acceleration. Notice we're not actually giving it numerically we're given in words but we know the final velocity. The final velocity is 0 because it decelerates to rest, rest means not moving, so that's a velocity of 0.

So we know initial and final velocity, we know a, we're looking for distance, notice what do we not know and not care about? We don't know time and we don't care about time. In fact this is classic for a braking problem. Whenever you brake, you want to know the distance, you want to know the initial speed, the final speed.

You're typically not interested, during a process of braking you're not interested in the time it takes to brake. You're interested in the stopping distance, the distance it takes to brake. So the time independent equation is often the one we use in a braking scenario. So here's the time independent equation. We'll plug in values.

We're looking for the final distance. We get 20 squared, of course that's 400. We will divide both sides by -4, and so we'll get positive 100. And the 2, the 100 divided by 2 is going to be 50, and then meters squared per second squared divided by meters per second squared.

The seconds will cancel and we'll just be left with meters. And so what we'll wind up with is 50 meters. So that is the distance covered during deceleration, half a football field, 50 meters. Notice, incidentally, that that was 20 miles an hour, 20 meters per second, about 40 miles an hour, -4 meters per second, that's an awfully realistic deceleration for a car with very good brakes.

So notice you're going 40 miles an hour, that's a stopping distance of a half a football field, that's very important to appreciate. So again, answer here, answer choice B. Now we'll talk about freefall. If we can ignore air friction, anything falling near the earth's surface accelerates downward in earth's gravitational field.

And the acceleration has this special letter g, g is the acceleration due to free fall. The technical value is 9.8 meters per second squared, and we often approximate this without a calculator. Often, this is approximated as 10 meters per second squared. Many of that problems that we'll have over the next few units, I'll be using ten meters per second squared as the approximation value.

You should know that technically, it's 9.8. So sometimes you approximate with ten then you have to realize that you have to estimate that the value is slightly less. This is the downward acceleration for anything dropped or thrown up. Something thrown up in the air decelerates on the way up and then it speeds up on the way down.

And so, notice if I throw a ball in the air, I throw it up and it goes up and comes back down, that entire process Is freefall. Both the part where it's going up, the stopping at the top, and the coming down. All of that is freefall, because that entire time it is under the influence of nothing but gravity. One big idea of freefall, Galileo's big discovery, is that freefall acceleration doesn't depend on mass.

Heavy things and light things fall at the same rate. So, of course, the legend that we have is that he went to the top of the Leaning Tower of Pisa and dropped a big rock and a small rock. Probably that didn't happen, but we do know that he did lots of experiments rolling balls down ramps, this sort of thing. And he definitively measured that, at least over small distances, two objects will fall at the same rate regardless of their weights.

That's very, very important. So there are two caveats to this general pattern. If we're talking about extremely light objects, feathers, pieces of paper, dust particles, then air friction is significant. So that doesn't count as freefall. Any time that you get to a point that the air friction is taking over, that's no longer freefall.

If I drop a brick from my hand, air friction is negligible, so I can count that as freefall. If I drop a feather from my hand, air friction takes over right away, so that doesn't count as freefall. Also, over large enough distances, over a large enough drop, any object will reach what's called terminal velocity.

And again, if I'm dropping something from my hand, a brick is not gonna reach terminal velocity. If I drop a brick out of an airplane, now I'm not recommending that you do this for safety reasons. But if you drop a brick out of an airplane, over that long distance the brick eventually will reach something called terminal velocity.

It will be moving so fast that the air friction will be balancing the weight of the object, and it will be actually falling at a constant velocity. We'll discuss terminal velocity more in the next module when we discuss forces, so right now we're gonna ignore these two cases. We're just gonna talk about short distance dropping, and we'll assume that air friction is negligible.

We also have to be careful with plus and minus signs, especially when some of the motion is up and some of the motion is down. It actually doesn't matter at all whether we say down is positive or down is negative. We can make either choice, the important thing is that we have to be consistent. We have to make sure that every value, the distance, the speed, and g, the acceleration, that they all follow the same convention.

As long as we're consistent in our convention, it doesn't matter whether we say up is positive or down is positive. We can use any of the kinematic equations, so remember those four kinematic equations. We can use those for something in freefall, and we always use g for the value of the acceleration.

For example, if I drop something from rest down a well, and it hits the water 5 seconds later, how deep is the water level in that well? Well, I have a time of 5 m/s. I have an acceleration of 10 m/s, I'll use that approximation. The initial velocity is 0. And again, I'm gonna approximate 10 m/s, rather than using 9.8.

And I'm also gonna say, since all of the motion is downward, we'll just say that down is positive for the sake of this problem. So that's why 10 is positive here. So we'll use this equation. This is the equation we use when we don't care about final velocity. Notice the v initial is 0, so that term goes away.

And it's just 1/2 acceleration times 5 squared, so that's 5 m/s squared times 25s squared. 5 times 25 is 125. The seconds squared cancel, so we get 125m. So that is the depth of the well. Incidentally, we also could figure out the final velocity if we wanted.

The final velocity we'd use this equation, 10 times 5, 50 m/s. So in other words, at the end of 5 seconds it's falling about 100 miles an hour. That's how fast it hits the water. So it's a little more than a football field in depth, and it hits that water very, very fast. Here's a practice problem.

Pause the video, and then we'll talk about this. Okay, this is a tricky problem. A lot is going on in this problem. A man at the edge of a canyon throws a stone straight up at 30 m/s. It goes up, then drops down past its initial level, all the way down to the floor of the canyon.

If the stone hits the canyon floor exactly 10 seconds after it left the man's hand, how high was the man above the canyon floor, and how fast did the stone hit the canyon floor? Wow, all right, a few things to notice. First of all, this person probably has a certain amount of athletic ability if they can chuck a stone up at 60 miles an hour.

That's pretty impressive. But okay, we'll just assume we have someone with a good arm. So how are we gonna approach this? I'm gonna say, let's consider the trip up from his hand to stopping. We're gonna call up negative and down positive, Okay?

And incidentally that convention, up is negative, down is positive, we're gonna follow that through the whole problem. So the initial velocity, it's an upward velocity, so that's negative -30 m/s. The acceleration is positive, and on the trip up the final velocity is 0, because it goes up and then it stops. So that counts as 0.

So we can figure out the time, and it turns out that it's 3 seconds, so it's 3 seconds on the way up. So that means this trip up is 3 s, and then the trip down is 7 s. When it goes up the average velocity is 15 m/s, and it goes up a distance of 45 m. So that is the height above the man's initial level.

That's the height that the stone reaches, and from there it falls. And so when it falls it starts at 0, it gets to a final velocity of 70 m/s. And incidentally that's part of the answer, that's how fast it hits the canyon floor. And we can figure out the distance it travels.

5*49, well let's think about this. Let's go off to the side here. We know that 5*50 = 250. 5*49 is, of course, 5(50-1). So that's gonna be 250-5, so that's gonna be 245. So that's the total distance, but that's the total distance from its stopping point way above where the man was, all the way down to the canyon floor.

And that distance above where the man was, that was 45 meters, so we have to add that. And so the actual distance from the man's level to the canyon floor is 200 meters. And notice that we get that answer. Another way to approach this problem incidentally, in fact a much quicker way, is to realize, again, 30 m/s, it's gonna take 3 seconds to go to rest.

Then it's gonna have 3 more seconds coming down to the man's level again, because going up and going down are symmetrical. So it's gonna come back down to the man's level in 3 more seconds, going 30 m/s. Then another 4 seconds it's gonna fall, it's gonna hit the ground at 70 m/s. So just think about the trip from the man's level going downward to the bottom of the canyon.

Start at 30, end at 70, average velocity 50. And if you fall at an average velocity of 50 for 4 m/s, that's 200m. That's a more elegant way to solve for the distance. In summary, we talked about acceleration. We talked about the four kinematic equations. All of these are true, incidentally, for constant acceleration.

If the acceleration is changing then we have to use calculus, it's a whole other ball game. Typically that does not show up on the MCAT. We talked about freefall, how it can be approximated by 10 m/s squared. We can use any of those kinematic equations, and we would just use this value for the acceleration.

Read full transcriptBut in Physics, acceleration means any change in speed, getting fast or getting slower. Those are two different kinds of acceleration. Sometimes we call the getting slower, deceleration. But that's just a kind of acceleration. We will consider only uniform acceleration.

That means that the acceleration itself is not changing. So, if the speed is changing, but the acceleration itself is staying constant. Now, what's tricky is that if acceleration starts changing, well then, all the equations have to involve Calculus, then we're doing Physics with Calculus. And Physics with Calculus is a little beyond what the MCAD is gonna ask of you. And so, on the MCAD, you will see constant acceleration.

You're not gonna see a whole lot of changing acceleration. The fundamental equation for constant acceleration is just acceleration equals change in speed over time. So V final minus V initial over t. And this can be rewritten in this form, so that is just a rewritten form of the definition of acceleration.

And this is one of the fundamental kinematic equations. So, store that in your memory banks, we'll go back to that, that's kinematic equation number one right there. V final equals at plus V initial. If distance is in meters and time is in seconds, then speed would be in meters per second.

So, meters per second is the fundamental metric unit of speed and incidentally I'll say, just for approximation purposes. If you take a speed in meter per second and double the number that's approximately the speed in miles per hour. So, something's going 30 meters per second, that's approximately 60 miles an hour, that's just a good approximation to get a ballpark idea about speed.

What are the unites of acceleration? Well, an acceleration is speed divided by time, so that's meters per second divided by second, and we often write that as meters per seconds squared. And that's something that can be confusing to people, what is does that mean? Meters per seconds squared. Well, example an acceleration of 3 meters per second squared means that for every second the speed is increasing by an additional 3 meters per second.

So, during the first second, let's say it starts out at rest. After one second it the object is going 3 meters a second. After two seconds it's going 6 meters a second. Then 9. Then 12. Then 15.

Then 18 meters per second. So that speed is climbing and that's exactly what the acceleration of 3 meters per second means. One way to think about this stuff is graphically. Uniform acceleration, that is acceleration, constant acceleration. Acceleration, a constant value, appears as a straight line on a graph of speed versus time.

And so, you can see what's going on here, the slope is the acceleration. And so what's happening is with this line, of course, the speed is going up and up and up, but the slope is remaining constant. That is the constant acceleration. Here, the slope is going down, so the speed is going down, down, down but again the slope is constant.

That's the constant acceleration. Notice this is on a speed versus time graph. Now, instead if we have a distance versus time graph, an x versus t graph, then acceleration appears as a parabola. On this graph, the slope is the speed. So notice what's happening.

Here, we're going from low slopes to steeper and steeper slopes. The graph is getting steeper and steeper. So that means the speed is increasing. Here what's happened is the slope starts out steep and then it gets less and less steep. So, here we have deceleration.

So, acceleration is an upward curving parabola, deceleration is a downward curving parabola. At least when it's moving up like this. Now constant speed appears as a straight line on as an x-vs-t graph. And appears as a horizontal line as a v-vs-t graph. So notice, it's very important.

You can't just relay on the shape to tell you what the motion is. You have to pay attention to whether it's a speed versus time graph or a distance versus time graph. So these two graphs here, these are shown exactly the same motion, constant velocity motion, but on two different graphs. On the v versus t graph, because the speed is constant, it's just a horizontal line.

As time goes on, it's staying at that one speed. Whereas on the x versus t graph, obviously position is changing, but the slope, the speed, is remaining constant. Again, you can't just rely on shape, you have to pay attention to whether it's a v versus t or an x versus t graph. So, here's a practice problem.

Pause the video, and then we'll talk about this. Okay, let positive x indicate the direction a car is facing. Which of the following distance vs time graphs would represent a car in reverse accelerating backwards? So this is the driver who puts the car in reverse and then steps on the gas, so the car accelerates backwards.

Well, let's see. What's going on in A? So notice, first of all, any of the graphs where it's going from down to up, the car is moving forward. So A, what's happening is it's moving forward, and we also have slope increasing.

So this is accelerating forward. A would be foot on the gas, car in drive. Notice that C is also going up, but here we have deceleration. We go from high slope to a lower slope. So this would be car on drive, but foot on the brake. Decelerating forward.

So, it's gonna be either B or D cuz both of these are in drive, they're not in reverse. We look at B, what's happening is, we have a big negative slope, a negative slope that is very steep, and it's getting less and less steep in going to zero. So in other words this is deceleration backwards, this is car in reverse, put on the brake.

Where as in D, we start out with a low slope and then the slope gets more and more negative. So, it's going more and more in the backwards direction. So this is the car in reverse, foot on the gas and this is the one we want, car in reverse accelerating backwards, answer choice D. Now you may remember from the last video, if we have two different stretches of constant velocities at two different speeds, we can't simply average the two speeds to get the average velocity.

And so, if you remember the practice problem from last time, this is actually a graph of that practice problem. For six hours, going at 20 miles an hour, and then for two hours going at 60 miles an hour. And of course, what we have to do is figure out the individual distances, individual times, in order to figure out the average velocity.

We did that in the last lesson. By contrast, when a speed uniformly increases over the interval, when you have constant acceleration. We can numerically average the individual final velocities. So here we have a v versus t graph of something that is accelerating. It starts at 30 meters per second, and then over the course of eight seconds, it actually accelerates to 70 meters a second.

And so this is the basic setup, from this notice we could figure out the acceleration. That would be change in velocity divided by time. Five meters per second squared. That's a pretty good acceleration right there. If you could get a car that accelerates at five meters per second squared that's an awfully decent acceleration.

Notice also, we can average the beginning and the ending velocities to get the average velocity. The average velocity of this entire trip is going to be 50 meters a second. And so this is a very different case. If we have two plateaus as we saw in the previous, we can't just average those velocities.

But if we have a constant smooth straight line, the average height of this straight line is the average of the starting point and the ending point. And that's why we can average the starting point and the ending point to get the average velocity. So far, we have two important kinematic equations. So the first one I re-wrote it a bit, v final minus v initial equals AT.

The second one, average velocity equation, total distance divided by time equals v initial plus v final divided by two. Now notice we can multiply these to cancel t. So I'm gonna multiply them like this. Multiply the AT times x total divided by t and multiply that difference, v final minus v initial by that fraction.

What happens, of course, the t's cancel and on the left we're gonna multiply both sides by 2, just to move that 2 in the denominator to the other side. And we get the difference of two squares, that algebraic pattern a minus b times a plus b equals a squared minus b squared. And that gives us this equation. That's a really important equation.

Sometimes this is called the time independent equation because this equation has all the other kinematic factors but it doesn't have time. So that is the third important kinematic equation, again, called the time independent equation. Start again with those two equations. And what we're gonna do this time is plug the left equation into the right equation.

And so, we'll plug in like this, we get 2 of the v initials. Then what we're gonna do is separate out the fraction, so we get 1/2 at + v initial, and then just multiple both sides by t. So we get X total equals one half AT squared plus ViT. That's the fourth big kinematic equation.

And so these four equations are four equations you need to know. And so here they are again. And notice what's important about each equation is not only noticing what values are in each equation, but also being aware of what is not in each equation. Each one of these is missing a different thing. And so, there's a problem where we don't care about acceleration, we might use equation 2.

If we don't know and don't care about the final velocity, we might use equation 4. And so it's very important to be aware of not only what each equation has, but also what it doesn't have. Part of the basic strategy in any kinematic problem is to discern which equation to use from what you're given, what you need to find, and also what you are not given and what you don't care about.

All of those together go into the choice of the right kinematic equation. Here's a practice problem. Pause the video, and then we'll talk about this. Okay. A car, initially traveling at 20 meters per second, decelerates to rest at negative four meters per second squared.

How much distance is covered during this deceleration? So a few things. Notice we know initial velocity, we know an acceleration. Notice we're not actually giving it numerically we're given in words but we know the final velocity. The final velocity is 0 because it decelerates to rest, rest means not moving, so that's a velocity of 0.

So we know initial and final velocity, we know a, we're looking for distance, notice what do we not know and not care about? We don't know time and we don't care about time. In fact this is classic for a braking problem. Whenever you brake, you want to know the distance, you want to know the initial speed, the final speed.

You're typically not interested, during a process of braking you're not interested in the time it takes to brake. You're interested in the stopping distance, the distance it takes to brake. So the time independent equation is often the one we use in a braking scenario. So here's the time independent equation. We'll plug in values.

We're looking for the final distance. We get 20 squared, of course that's 400. We will divide both sides by -4, and so we'll get positive 100. And the 2, the 100 divided by 2 is going to be 50, and then meters squared per second squared divided by meters per second squared.

The seconds will cancel and we'll just be left with meters. And so what we'll wind up with is 50 meters. So that is the distance covered during deceleration, half a football field, 50 meters. Notice, incidentally, that that was 20 miles an hour, 20 meters per second, about 40 miles an hour, -4 meters per second, that's an awfully realistic deceleration for a car with very good brakes.

So notice you're going 40 miles an hour, that's a stopping distance of a half a football field, that's very important to appreciate. So again, answer here, answer choice B. Now we'll talk about freefall. If we can ignore air friction, anything falling near the earth's surface accelerates downward in earth's gravitational field.

And the acceleration has this special letter g, g is the acceleration due to free fall. The technical value is 9.8 meters per second squared, and we often approximate this without a calculator. Often, this is approximated as 10 meters per second squared. Many of that problems that we'll have over the next few units, I'll be using ten meters per second squared as the approximation value.

You should know that technically, it's 9.8. So sometimes you approximate with ten then you have to realize that you have to estimate that the value is slightly less. This is the downward acceleration for anything dropped or thrown up. Something thrown up in the air decelerates on the way up and then it speeds up on the way down.

And so, notice if I throw a ball in the air, I throw it up and it goes up and comes back down, that entire process Is freefall. Both the part where it's going up, the stopping at the top, and the coming down. All of that is freefall, because that entire time it is under the influence of nothing but gravity. One big idea of freefall, Galileo's big discovery, is that freefall acceleration doesn't depend on mass.

Heavy things and light things fall at the same rate. So, of course, the legend that we have is that he went to the top of the Leaning Tower of Pisa and dropped a big rock and a small rock. Probably that didn't happen, but we do know that he did lots of experiments rolling balls down ramps, this sort of thing. And he definitively measured that, at least over small distances, two objects will fall at the same rate regardless of their weights.

That's very, very important. So there are two caveats to this general pattern. If we're talking about extremely light objects, feathers, pieces of paper, dust particles, then air friction is significant. So that doesn't count as freefall. Any time that you get to a point that the air friction is taking over, that's no longer freefall.

If I drop a brick from my hand, air friction is negligible, so I can count that as freefall. If I drop a feather from my hand, air friction takes over right away, so that doesn't count as freefall. Also, over large enough distances, over a large enough drop, any object will reach what's called terminal velocity.

And again, if I'm dropping something from my hand, a brick is not gonna reach terminal velocity. If I drop a brick out of an airplane, now I'm not recommending that you do this for safety reasons. But if you drop a brick out of an airplane, over that long distance the brick eventually will reach something called terminal velocity.

It will be moving so fast that the air friction will be balancing the weight of the object, and it will be actually falling at a constant velocity. We'll discuss terminal velocity more in the next module when we discuss forces, so right now we're gonna ignore these two cases. We're just gonna talk about short distance dropping, and we'll assume that air friction is negligible.

We also have to be careful with plus and minus signs, especially when some of the motion is up and some of the motion is down. It actually doesn't matter at all whether we say down is positive or down is negative. We can make either choice, the important thing is that we have to be consistent. We have to make sure that every value, the distance, the speed, and g, the acceleration, that they all follow the same convention.

As long as we're consistent in our convention, it doesn't matter whether we say up is positive or down is positive. We can use any of the kinematic equations, so remember those four kinematic equations. We can use those for something in freefall, and we always use g for the value of the acceleration.

For example, if I drop something from rest down a well, and it hits the water 5 seconds later, how deep is the water level in that well? Well, I have a time of 5 m/s. I have an acceleration of 10 m/s, I'll use that approximation. The initial velocity is 0. And again, I'm gonna approximate 10 m/s, rather than using 9.8.

And I'm also gonna say, since all of the motion is downward, we'll just say that down is positive for the sake of this problem. So that's why 10 is positive here. So we'll use this equation. This is the equation we use when we don't care about final velocity. Notice the v initial is 0, so that term goes away.

And it's just 1/2 acceleration times 5 squared, so that's 5 m/s squared times 25s squared. 5 times 25 is 125. The seconds squared cancel, so we get 125m. So that is the depth of the well. Incidentally, we also could figure out the final velocity if we wanted.

The final velocity we'd use this equation, 10 times 5, 50 m/s. So in other words, at the end of 5 seconds it's falling about 100 miles an hour. That's how fast it hits the water. So it's a little more than a football field in depth, and it hits that water very, very fast. Here's a practice problem.

Pause the video, and then we'll talk about this. Okay, this is a tricky problem. A lot is going on in this problem. A man at the edge of a canyon throws a stone straight up at 30 m/s. It goes up, then drops down past its initial level, all the way down to the floor of the canyon.

If the stone hits the canyon floor exactly 10 seconds after it left the man's hand, how high was the man above the canyon floor, and how fast did the stone hit the canyon floor? Wow, all right, a few things to notice. First of all, this person probably has a certain amount of athletic ability if they can chuck a stone up at 60 miles an hour.

That's pretty impressive. But okay, we'll just assume we have someone with a good arm. So how are we gonna approach this? I'm gonna say, let's consider the trip up from his hand to stopping. We're gonna call up negative and down positive, Okay?

And incidentally that convention, up is negative, down is positive, we're gonna follow that through the whole problem. So the initial velocity, it's an upward velocity, so that's negative -30 m/s. The acceleration is positive, and on the trip up the final velocity is 0, because it goes up and then it stops. So that counts as 0.

So we can figure out the time, and it turns out that it's 3 seconds, so it's 3 seconds on the way up. So that means this trip up is 3 s, and then the trip down is 7 s. When it goes up the average velocity is 15 m/s, and it goes up a distance of 45 m. So that is the height above the man's initial level.

That's the height that the stone reaches, and from there it falls. And so when it falls it starts at 0, it gets to a final velocity of 70 m/s. And incidentally that's part of the answer, that's how fast it hits the canyon floor. And we can figure out the distance it travels.

5*49, well let's think about this. Let's go off to the side here. We know that 5*50 = 250. 5*49 is, of course, 5(50-1). So that's gonna be 250-5, so that's gonna be 245. So that's the total distance, but that's the total distance from its stopping point way above where the man was, all the way down to the canyon floor.

And that distance above where the man was, that was 45 meters, so we have to add that. And so the actual distance from the man's level to the canyon floor is 200 meters. And notice that we get that answer. Another way to approach this problem incidentally, in fact a much quicker way, is to realize, again, 30 m/s, it's gonna take 3 seconds to go to rest.

Then it's gonna have 3 more seconds coming down to the man's level again, because going up and going down are symmetrical. So it's gonna come back down to the man's level in 3 more seconds, going 30 m/s. Then another 4 seconds it's gonna fall, it's gonna hit the ground at 70 m/s. So just think about the trip from the man's level going downward to the bottom of the canyon.

Start at 30, end at 70, average velocity 50. And if you fall at an average velocity of 50 for 4 m/s, that's 200m. That's a more elegant way to solve for the distance. In summary, we talked about acceleration. We talked about the four kinematic equations. All of these are true, incidentally, for constant acceleration.

If the acceleration is changing then we have to use calculus, it's a whole other ball game. Typically that does not show up on the MCAT. We talked about freefall, how it can be approximated by 10 m/s squared. We can use any of those kinematic equations, and we would just use this value for the acceleration.