What Is Acceleration? Why It's 180, Not 20

Cold open: more points is not faster progress

0:00
since efficiency is what counts. I'm more efficient. If I worked a bit harder like her, I'd get all 110 points, right? Clearly I'm better. So note: even with a bigger change, efficiency isn't necessarily higher. Progress isn't necessarily faster. It may just come from piling on hours.

Why this lesson: the follow-up to Δv

0:25
This lesson is on acceleration, the most important in describing motion. Acceleration is easily a top-three quantity in high school physics. Last lesson, on change in velocity, was just a setup for acceleration. I want you, after these two lessons, to truly understand acceleration. Once you truly get it, it helps a lot in later physics. So I strongly recommend: if you haven't seen change in velocity, watch that lesson before this one. Watch the two lessons back to back, you'll fully get it.

Two students, one question: who did better?

1:14
First, a question for you. Here are Qi Xiaoyu and me in high school, ours was the old college entrance exam, with physics out of 110 points. My physics score went from 70 to 80. Hers went from 60 to 90. Which of us is better? If you pick me, type 1. If you pick Qi Xiaoyu, type 2. I bet more of you typed 2. Because she improved more. She gained 30 points, I gained 10. Her improvement is larger. Isn't that the change we covered last lesson? Her change is +30 points. Mine is +10. Her change is bigger. But a bigger change doesn't mean higher study efficiency. Maybe she just brute-forced it with time. A semester is about 4 months, 120 days.

2:18
Qi Xiaoyu studied nonstop, really: 120 days, not one day off. She studied seriously for all 120 days and went from 60 to 90. She studied 120 days. And me? I thought physics was easy, so I played and studied only the last 10 days before the exam. I studied just 10 days. Yet I gained 10 points. Looking at it that way, who's better? Now I bet more of you type 1. You'd say I'm better, since efficiency is what counts. I'm more efficient. If I worked a bit harder like her, I'd get all 110 points, right? Clearly I'm better. So note: even with a bigger change, efficiency isn't necessarily higher. Progress isn't necessarily faster. It may just come from piling on hours.

3:23
Suppose I want to study whose efficiency is higher, who improves faster. For such questions, if I give only the change, not the time, you can't figure it out, right? So for such questions, change alone isn't enough. So what now? Here's what we can do. Many of you already feel this: Since I gained 10 points in 10 days, I divide 10 points by 10 days to get the average gain per day. Qi Xiaoyu: 30 points in 120 days, so I divide 30 points by 120 days. Dividing gives the average number of points gained per day. I divide the change by the time interval. It's like fixing the time. Dividing both, I take one day for each and compare points gained per day. The more points gained per day, the higher your efficiency, right?

4:31
Doesn't this division show which of us improved faster, and who studies more efficiently? It measures how fast things change. Make sense? Divide the change by the time interval, and you see how fast it changed. How fast things change is our focus today: rate of change.

Rate of change: change per unit time

5:00
How fast it changes: the rate of change. In full, it's the change per unit time. Like my division on the last slide, where one day is the unit time. Score change per day is my rate of change. Change per unit time is rate of change. How long a unit time to use depends on the problem. In astronomy, for example, a unit of one year is common. And for exam points, I might use days or weeks, or months. Using seconds would be silly, Score change per second? Weird. But in high school physics, we usually take one second. Because we study motion problems, and the second is a standard unit. So we use one second as the unit time. Rate of change equals the change divided by the time interval.

6:08
That's the quantity we just computed. Top is the change; bottom is the time interval over which it happened. Be sure to tell change apart from rate of change. Note: change shows how much it changed. How many points you gained: 10, 20 or 50? That's how much. Rate of change shows how fast it changes. That's the average points per day. Points gained per day is how fast it changes. Many physical quantities are defined by rate of change, like velocity, which we just learned. Velocity is the rate of change of displacement. What's displacement? The change in position. Say the start is A and the end is B. Then the position change from A to B is described by displacement.

7:15
Bigger displacement, more position change. Say there's a point C. Displacement AC is bigger than displacement AB. From A to C, position changes more than from A to B. Displacement: how much position changes. Velocity shows how fast position changes. Higher velocity: position changes faster. Two people both go from A to B. The first takes 1 s, and the second takes 100 s. Say A and B are 5 m apart, so the displacement is 5 m. Work out their velocities. The first person's is bigger. The second person's is smaller. The first goes A to B in 1 s, the second takes 100 s to slowly walk from A to B. Who changes position faster? Clearly the first one.

8:17
How fast position changes is the position change per second. So velocity is the rate of change of displacement.

Large change vs large rate of change

8:26
About change and rate of change, I want you to master these two statements. Once you get them, you're all set. Statement 1: a big change means a lot changed, but not necessarily fast. Change describes how much changed. If it's large, much has changed. But not necessarily fast. How do we define fast or slow? We use the average per second, or the change per unit time. Divide by the time interval, maybe huge, and the value gets small. Take Qi Xiaoyu's score: a big jump, but over 120 days. Divide, and her rate of change is slow. Statement 2: a large rate of change means it changes fast. That's certain, because rate of change describes how fast. So a large rate means fast change.

9:30
But not necessarily a lot. Yes, you change fast: a lot per second on average. But maybe it changes for only a short time. To find the change from a rate, multiply by time. If the time is short, the change needn't be large. From these two statements, you should now see change vs. rate clearly. Now let's learn the rate of change of velocity: acceleration.

Acceleration is the rate of change of velocity

10:04
Acceleration measures how fast velocity changes. It's the rate of change of velocity. But it's used so often, "Rate of change of velocity" is a mouthful, so we gave it a name: acceleration. Also, acceleration and rate of change of velocity are always exactly equivalent. Acceleration: rate of change of velocity. And the reverse is true too. That's how the two relate: same thing. It shows how fast velocity changes. Larger acceleration means faster change. In daily life, we feel acceleration a lot. We're sensitive to how fast speed changes. Say you're in a car. If it drives forward smoothly, constant speed, velocity unchanged, acceleration is zero.

11:08
You feel nothing; can't tell it's moving. Like a smooth high-speed train. You read on the train and feel fine. Same as sitting at home. Because a is zero, speed is steady; you feel nothing. But if you're riding in a car with a reckless young driver, who doesn't press the gas smoothly, he floors it, and the car surges. With such large acceleration, your heart nearly jumps out. Quite a feeling! Or a car that's fast brakes suddenly, and you feel awful. When acceleration is large, velocity changes fast, people feel uncomfortable. Why do some people get carsick? One reason: the driver isn't experienced. Starting and braking, he can't press smoothly. His acceleration is often large. Fast start, then fast brake, again and again. You'll feel sick. This is how acceleration affects us.

a = Δv/Δt and the unit m/s²

12:13
Acceleration is defined as a = Δv / Δt, the rate-of-change formula from earlier. Acceleration is a rate of change. Over a given time, take the change in velocity, divide by time, and you get the acceleration. Note: by this definition, this acceleration is for that interval's average rate of change. Or see it as the average change in velocity per second. For example, a car speeds up to the right, unevenly. First second: velocity up by 2 m/s. Next second, up by 10 m/s. Third second, up by 1 m/s. Total change in velocity over 3 s? Let's add them up. 13 m/s. So how do I find a? Divide the 13 m/s change by 3 s. That gives the average change in velocity per second.

13:22
Note this definition gives an average. If velocity changes unevenly, you get the average. Its unit is meters per second squared. Where does it come from? Look: Δv has units of meters per second, subtracting one velocity from another gives velocity units. So Δv is in m/s. Then divide by time, whose SI unit is s, giving m over s squared. So there's a square, read "meters per second squared." One last reminder. Beginners, take note: don't be fooled by the name. Many think, with that name, it only means speeding up. Not so! It's how fast velocity changes. Two cases: If the car is speeding up, it's how fast velocity increases.

14:29
If it's slowing down, it's how fast velocity decreases. So it covers both speeding up and slowing down. A change can be an increase or a decrease. So don't be fooled by the name.

Practice: two kinds of "fast"

14:48
Now two problems to reinforce the concept. Problem 1, from a school exam, Zhejiang 9+1 Alliance, 2019-2020. Xiao Zhang watches an F1 race and sees a car flash by, and says, "So fast!" At the finish the car brakes to a stop, Xiao Zhang says "So fast!" again. Which statement is correct? In daily life, "fast" can sometimes mean that it moves fast, i.e. high speed, or that velocity changes fast. Everyday speech is loose. So what's the first "fast"? The car isn't necessarily speeding up. It may move at a constant, high speed. It whips past you in an instant. That means its speed is high, so the first "fast" means the car's speed is high. It's gone before you see it, it's just too fast.

15:53
And the second? The car braked suddenly and stopped. Wasn't that a quick brake? From high speed, stopped at once. Wow, what a fast stop! So the second "fast" should mean that velocity changes fast. The answer is B. A great problem; it links what we just learned to real life. And real life in turn helps us understand physics better. So the answer is B. Next, another one.

Practice: four statements about acceleration

16:28
This problem is from Changjun High School. Let's take a look. Here are a few situations. 1: A car turns at an intersection. Its speedometer reading stays constant. 2: A rocket just ignited, about to lift off. 3: A truck's emergency braking on a highway. 4: A maglev train running at high speed. Which judgment below is correct? A: The speedometer reading is constant, so the car's velocity is constant. Careful, that's wrong. That number on the dashboard shows only the magnitude of velocity, which is speed. But velocity is a vector, with magnitude and direction. When a car turns, its direction is surely changing.

17:34
For a vector like velocity, either one changing means it changes. Only if both are fixed is it constant. A turning car surely changes direction, so its velocity changes when it turns. If we change "velocity" to "speed," it's right. Constant reading means constant speed. Straight or turning, the magnitude is unchanged. So when a car turns, its velocity changes, but speed doesn't. So A is out. B: It hasn't moved, so a must be zero. To tell if acceleration is zero, look at initial and final velocity. If both are zero, velocity didn't change then, so a is surely zero. But this rocket hasn't moved yet; only the initial velocity is zero. After 0.1 s, the final velocity may become 100 m/s.

18:37
Then a = (100 - 0) / 0.1 is definitely not zero. You can't say a is zero just because the rocket isn't moving yet. B is wrong. C: When a truck brakes hard, change in velocity is large, and so is a. Let's analyze. The truck is moving at high speed. What does braking mean? Speed drops to zero. That's braking. From a very high speed down to zero, the change in velocity is large here. So "change in velocity is large" is true. But is the acceleration also large? Recall my earlier slide: A large change means a lot of change, but not necessarily fast change. A large change doesn't mean a large rate. Why is that? There's a case: the change is large, but it takes a very long time.

19:43
Dividing by time gives a small number. So when the time taken is very long, a large change, a small rate, that is, slow change. But look at the situation here: it says emergency braking. What is emergency braking? The time taken is extremely short. Take a very large change in velocity and divide by an extremely short time: large. So the rate of change in velocity here, i.e. acceleration, is large too. So choose C. An extremely short time interval and a very large change in velocity, like a race car flooring the pedal to launch suddenly, or a car braking hard. Then the change in velocity is large, and so is acceleration. So C is correct. D: A maglev train moving at high speed Its speed is high, so a must be large.

20:50
Careful, it's wrong. A high-speed maglev train is fast, but usually at constant speed. Large acceleration is uncomfortable. It keeps a constant speed. So it moves fast at a constant speed, and its acceleration is zero, since velocity is constant. So "high speed means large a" is wrong. D is wrong; only C is correct. Great problem, worthy of a top school.

Acceleration is a vector: splitting the arrow

21:26
Now let's study acceleration itself. Does it have direction? Is it a scalar or a vector? Let me give an example. A cart has an initial velocity of 1 m/s, to the right. After accelerating for 3 seconds, its velocity becomes 10 m/s to the right. Take right as positive. First, the change in velocity over 3 s. 10 - 1 = 9 m/s and it points right. That's the change in velocity. Didn't I make it clear last lesson? Change in velocity has a direction, It's an arrow. Let me draw this 9 m/s as an arrow. Say the arrow points right, with length 9. This arrow is the change in velocity. Now think: what is acceleration?

22:33
Acceleration takes a change in velocity and averages it over each second, because you divide by the time. It's the average change in velocity per second. So the essence of acceleration is also a change in velocity per second. Since the whole change in velocity is an arrow, my acceleration comes from splitting it into three equal parts. One arrow becomes three arrows. Each of these arrows is my acceleration. So acceleration is also an arrow, so it has a direction. I split one full change in velocity, one whole arrow, into equal parts, giving several arrows. Each arrow is cut from the big arrow, so each one points the same way as it. So acceleration is a vector, with direction.

23:38
a points in the same direction as Δv. They're essentially the same thing. Only the time you take is different. Whole interval: change in velocity. Average it over each second, and you get acceleration. Got a change in velocity? Split it into equal small arrows by time. So they point the same way. Acceleration is a vector with direction. With this drawing, the direction of a and Δv is obvious, right? Just by splitting the arrow. I bet only I teach it this way. Others may skim past it, but I think it's worth explaining well. One piece of advice: for now, don't dig into what its direction means. I know many of you are curious. What does its direction mean? Its direction is very important, but we're not ready for that yet.

24:43
When we get to Newton's second law, I'll tell you exactly what it means. Let's set this aside for now. Not skipped, just not the right time yet. OK, that's enough on direction. One last question on this page: How do we find acceleration in straight-line motion?

Computing acceleration on a line

25:02
Actually it's simple. Acceleration is just change in velocity divided by time, right? The time is usually given. So the core of finding acceleration is last lesson's change in velocity. That's the hard part and the core. Step 1: choose the positive direction. Same as the steps for change in velocity. Sort out the signs of v₀ and vₜ. Step 2: calculate Δv. Step 3: divide by the time. Done. That simple. Let's practice with a few problems.

Example: a ball bouncing off a wall

25:47
A ball hits a wall horizontally at 10 m/s. Within 0.1 s it bounces back at 8 m/s, same straight line. Find the acceleration during the collision That is, during this 0.1 s. How to solve it? Same steps. First, the positive direction. Take the initial velocity's direction. The ball flies toward the wall, perpendicular to the wall, inward. So inward is the positive direction. My v₀ is +10 m/s, and my vₜ is −8 m/s. So Δv = −8 − 10 = −18 m/s. Then a = Δv / Δt = −18 / 0.1 = −180 m/s² So its magnitude is 180 m/s².

27:09
I took inward as positive, and a is negative, so it points opposite: outward. So the magnitude is 180, perpendicular to the wall, outward. So the answer is D. OK, no more on this one.

Only speeds given: two possible answers

27:26
Now the last problem of this lesson. An object's initial velocity is 6 m/s. After 10 s, the magnitude of its velocity is 20 m/s. How large could the acceleration be? Like a problem from last lesson. It has multiple possibilities because we only know the magnitude, not the direction. My approach is the same as last time: First choose a positive direction. I'll take the initial velocity's direction as the positive direction. So v₀ = 6 m/s, right? And the final velocity? I only know its magnitude is 20, but not whether it's + or −. So the final velocity could be +20 m/s or −20 m/s. Both are possible. So we must calculate + and − separately.

28:30
Two cases. Case 1: +20 m/s. Δv = 20 − 6 = 14 m/s Then a = 14 divided by 10 s = 1.4 m/s² Direction: positive. So that matches option B. It only asks for magnitude, so ignore the sign: just 1.4. Case 2: the final velocity is −20 m/s. Then Δv = −20 − 6 = −26 m/s Whatever the initial velocity's direction, I take it as the positive direction, so v₀ is always positive, so final minus initial: just subtract 6. So a = −26 / 10 = −2.6 m/s² Magnitude: 2.6.

29:35
So that's D. So this problem: B and D. Finding acceleration from its definition is fairly easy; two problems will do.

Summary

29:46
Two key points this lesson. First, what is acceleration? It is the rate of change of velocity, showing how fast velocity changes. whether it increases or decreases. Definition: a = Δv / Δt. Its direction matches Δv. Given straight-line motion, how do you find acceleration? The core of finding a is finding Δv, since after that you just divide by time. Step 1: choose positive direction. Step 2: calculate Δv. Step 3: compute a = Δv / Δt. Next: speeding up vs. slowing down.