Speeding Up or Slowing Down? Check a and v

Cold open: a shrinking, yet still speeding up

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doesn't affect our judgment. If a and v point the same way, the object is speeding up, even if a is getting smaller, its velocity still increases, just more and more slowly.

When does one quantity depend on another?

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Today: speeding up or slowing down. This topic really goes with acceleration, but that was too much for one lesson, so I split it in two. It's how we tell whether an object is currently speeding up or slowing down. A must-have skill in high school physics and a major use of acceleration. We'll nail this down in this lesson. First, I want to cover what velocity depends on. After acceleration, problems often ask Does v depend on a? Does a depend on v? and we can't analyze it. This point is important. It helps us see how physics defines the relationship between quantities. Let's clear this up. Let me be precise here. In physics, how do we define what depends on what, and analyze it?

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I depend on whatever can affect me. So what does "affect" mean? Holding all other conditions fixed, if you change, I must change with you, then I depend on you. What do grades depend on? On effort, right? With all else equal, the harder you work, the better the grades, and vice versa. Effort directly affects grades, right? So grades depend on effort. That's how we define "depends on."

What v, Δv and a each depend on

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Now, what does velocity depend on? That is, what factors can affect my v? An example will make it clear. Think of it as how much money I have, my net worth. What does my net worth depend on? First, on how much I had at birth. Say I was born into a rich family, I might do nothing and start rich. Born poor, I have no money. That matters a lot. It depends on what I started with. That's our v₀. v₀ is the initial velocity. Depends on it. Also on how much I've earned, from birth until now: my change in money. Say I earn a lot, then I get richer, right? If I'm a big spender, I might end up poorer. So it depends on the change in velocity We usually write v = v₀ + Δv.

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These two factors decide the velocity v. So I can say v depends on v₀, and v depends on Δv. Note what this arrow means: If v depends on v₀, I point from v to v₀. This arrow means v depends on v₀. v also depends on Δv, since both of them can affect v. Got it? Next: what does Δv depend on? Two factors: one is acceleration, the other is the time. Say the time is fixed. Say velocity changes for just 1 s, I speed up from zero. If my acceleration is 10 m/s², then Δv is 10 m/s: from 0 to 10. If my acceleration is 20 m/s², then in that one second, Δv is 20 m/s. With time fixed, bigger a gives bigger Δv.

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And with a fixed, the longer the time, the bigger Δv, right? Say a is fixed at 10 m/s² If the time is 1 s, then Δv should be 10 m/s. If I stretch the time to 2 s, Δv becomes 20 m/s. So a and t together decide Δv. So Δv depends on a, and also on t. Last: what does acceleration depend on? That's for a later lesson. Only two factors can affect acceleration: one is the net force F on the object, the other is its mass. Acceleration depends only on these two, and on nothing else at all. With the net force on an object fixed, more mass means smaller a.

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Mass fixed: more net force, larger a. It depends on these two things. Note: these relations go one way only. One is the cause, the other the effect.

Cause and effect: who influences whom

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It's very clear-cut. For example, v depends on v₀. The initial velocity affects v, right? So if v changes, v₀ is the cause, and v is the effect. It's the effect. Causes can affect effects, but not the other way around. If v₀ changes, v changes. But if v changes, does v₀ change? No, it doesn't. Because v₀ is from an earlier instant, and my current v is from a later one. The past can affect the future, but not the other way around, right? We all know that. This is physics, not a sci-fi movie. The future can't affect the past. So v₀ is the past, and v is the future relative to v₀, since it's the velocity at some instant after v₀. So I can only say v depends on v₀.

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You can't say v₀ depends on v, because v can't possibly affect v₀. Got it? v is the future. So if two physical quantities are related, one must be the cause and the other the effect. Be sure to tell which is which. One-way: never each other's cause. Same idea, right? Relative to v, Δv is also a cause. So you can say v depends on Δv, but not that Δv depends on v. Let me give you a perfect example This part is really fun. Here, v is like the son. v₀ is his mom, Δv is his dad, a is his grandpa, and t is his grandma. The son's height and build depend on Mom and Dad, right?

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His parents shape him. But whether Mom is pretty, Dad handsome, depends on me? Sounds weird, right? You can say I look like my dad, but "Dad looks like me" is weird, really weird. Or take me: I went to Peking University thanks to my grandpa. That sounds perfectly normal. since Grandpa went there. But could I say he went because of me? Would I dare say that? No way. When Grandpa went, I wasn't even born. How could I influence that? You can't say that. You know who's the son, who's the dad, who's the mom, grandpa, grandma. Now it's all sorted out. Now I'll ask you two questions. Your answers should be clear by now. Does v depend on a? Yes, it does.

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The son depends on grandpa. No problem: Grandpa influences him. Influence can pass down, right? No problem. So "v depends on a" is correct. But does a depend on v? Does acceleration depend on velocity? No, it doesn't. Grandpa doesn't depend on the son. Can the son affect his grandpa? No, it can't. So v depends on a: that's right. But a does not depend on v. Many students don't realize that, if two quantities have a relationship, there's always a cause and an effect. Many students don't know this, So "depends on" questions stump them. Once you see the cause and effect here, this problem is fully solved. Now let's sum up the relationship among a, v, and Δv.

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Three key lines. These three lines aren't to memorize, they're to understand.

Three conclusions about a, v and Δv

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First: v depends on a. Velocity depends on acceleration. Look at the diagram. But a alone can't determine v, because v depends on a, and also on v₀. Depending on a doesn't mean a alone decides v, unless v depends only on a. So v depends on a, but a alone can't determine v. You need v₀, and time t. Second: a does not depend on v or Δv. Note: a is the grandpa, v is the son, Δv is the dad. Son and dad can't affect grandpa. So grandpa a does not depend on Δv or v: Acceleration does not depend on either. Third: Δv depends on a. a can determine the direction of Δv: Δv points the same way as a.

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If acceleration points east, then after some time, Δv must point east. So a can determine the direction of Δv, but not its magnitude, since magnitude also depends on time t. So Δv depends on a: a sets direction, not magnitude. Once you understand these three lines, the relations are clear. Now, let's practice.

Practice: reading a = Δv/Δt correctly

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About a = Δv / Δt, which of these statements is correct? A: Δv / Δt is called the rate of change of velocity, meaning the magnitude of the change. Last lesson: how fast velocity changes, not how much. So A is wrong. B: The direction of Δv isn't always the same as a. We just covered this. a's direction decides Δv's direction, so they must be the same. So B is wrong. C: if velocity changes, Δv / Δt must be nonzero. Dividing by that time gives a nonzero result. Note: velocity can't jump instantly. A change in velocity takes time. So if velocity changes, Δv and Δt are nonzero, so Δv / Δt isn't 0. So C is correct. D: a is proportional to Δv and inversely to Δt.

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Note: proportionality is a relation. Direct proportion means: if I double, you double too, that's proportional. But we made this perfectly clear already: a doesn't depend on Δv or Δt. Only F and m set a. If a doesn't depend on Δv, Δt, can it be proportional? Or inversely proportional? First, there must be a relation, then ask if it's direct or inverse. No relation means no proportionality. So D is wrong. a doesn't depend on Δv or Δt at all, much less proportional. So the answer is C. Now let's formally cover speeding up vs. slowing down. First, review the lesson before last,

Review: direction of Δv and speed

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the one on change in velocity, where I explained what its direction means. Remember what I said? A change must show how much it changed, and whether it increased or decreased. For a scalar, a positive change means increase, negative means decrease. But for a vector, with direction, if the change has the same direction as the initial value, it's an increase. Opposite direction: a decrease. So if change in velocity points the same way as initial velocity, it must be speeding up: velocity rises. If it's opposite the initial velocity, velocity falls: it slows down. If you skipped the lesson before last, the one on change in velocity, go back and watch it.

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This is a full course: lessons connect. Watch them in order for best results. That's what the direction of Δv means: Same as initial velocity: speeding up; opposite: slowing down. Understand these two lines,

Same direction speeds up, opposite slows down

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and telling them apart is easy. Take v₀ as the velocity at one instant, and vₜ as the velocity at the next. If vₜ is greater than v₀, the object is speeding up, right? If vₜ is less than v₀, it's slowing down? So how do we tell which one? Easy. Just look at the directions of v₀ and a. We learned earlier that a and Δv always point the same way. If v₀ and a point the same way, then v₀ and Δv point the same way. By the line on the last page, it must be speeding up. So if a and v point the same way, it's speeding up, as in this figure. If v₀ points right, a does too, then Δv points right. Since vₜ = v₀ + Δv, together they make a longer arrow, so vₜ is longer: speeding up.

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Opposite a and v: this figure. Here, if a points left, then Δv points left too, since a decides Δv's direction. Then Δv is opposite to v₀, and they cancel. After cancelling, vₜ gets shorter, We covered this in detail in the change-in-velocity lesson. So I won't repeat it. Opposite a and v means slowing down. Here's the rule in short: Same speeds up, opposite slows. Remember, to tell them apart, v alone or a alone won't do. You need the directions of both v and a. Only the two together can decide. a and v same direction: speeds up. a and v opposite: slows down. To tell if it speeds up or slows, you must look at a and v. Only by looking at a and v can you tell if it speeds up or slows down.

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To judge, look at a and v together; a alone or v alone isn't enough. Rule: same speeds up, opposite slows. Memorize this rule. Same speeds up. a, v same: speeding up. a, v opposite: slowing down. Next, I've prepared tons of problems to master this. In your first term, it's key in school exams and tested often. So let's do several to master it. Let's start practicing.

Practice: a car's Δv and five statements

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A car moves rightward in a line, Original velocity: v₁, After a short time, it becomes v₂. Δv is the change in velocity. What follows from the figure? From now on, never use Δv to tell speeding up or slowing. Use a and v instead. Here Δv points left, so a must point left too, because a decides Δv, same direction. Now a and v₁ are opposite: v₁ points right, a points left. So it must be slowing down, a and v are opposite. So A is wrong. a's direction is opposite to v₁, so C is right and B is wrong. Last, D: a opposes Δv in direction. "Acceleration opposite to Δv" is wrong. Because the direction of Δv is decided by acceleration. Δv's direction? a decides it.

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So they must match. D is wrong too. The answer is C. No more on this one. Now, true-or-false items. First: acceleration is 0, so velocity must be 0? No. Zero acceleration means velocity is constant. Two cases. One: at rest, velocity 0. Two: uniform straight-line motion. Moving at constant velocity, velocity doesn't change either. So a = 0 has two cases. Second: velocity changes a lot, so acceleration is big? Not true. Big change doesn't mean big rate of change Covered last lesson. It depends on time too. Second: wrong. Third: velocity's rate of change is big, so acceleration's big? Yes. Acceleration is velocity's rate of change. Same thing, so this is definitely true. Fourth: velocity's direction is constant, and so is acceleration's direction. Careful: this is wrong. Let me give an example: a car.

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It keeps moving right, so v points right. First I hit the gas to speed up. While speeding up, a points right, right? After speeding up a while, I'm over the limit and must slow. So I ease onto the brake, but I don't fully stop, I keep slowing down. Now velocity is still to the right, but a points left. a and v opposite: slowing down. Look at v throughout the process: its direction never changed, always right. But a went right, then left: changed! Did a's direction stay the same? No, wrong. Last: Δv's direction is constant, so is acceleration? Careful: Δv's direction is always the same as acceleration's. So if Δv's direction is constant, acceleration's direction must be too. But its magnitude may change! You can't say a is unchanged: it's a vector, with magnitude and direction; it's constant only if both are.

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If Δv's direction is constant, we only know a's direction is constant. Magnitude: unknown. So the fifth is wrong too. Only the third one is right.

Practice: positive and negative accelerations

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Next: Object 1 and Object 2, moving along the same line. Object 1's acceleration is 3 m/s², Object 2's is -4. Which statement is correct? Note this one: we only know acceleration: magnitude and direction. Object 1 positive, Object 2 negative. Acceleration known, velocity unknown. Speeding up or slowing down? We can't. We need both a's and v's directions. Missing either won't work. "1 speeds up, 2 slows down" can't be determined. Motion direction unknown, so same or opposite direction? Unknown. But which changes velocity faster is known. Just compare magnitudes. Object 1: 3, Object 2: -4. Note: let me stress this again. When comparing two accelerations, the sign only shows direction. + means same as the positive direction - means opposite to the positive direction Minus doesn't mean below zero.

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So compare only the numbers, not the plus or minus signs. Object 2's number, 4, is bigger than 3, so Object 2's acceleration is greater. So Object 2's velocity changes faster. Bigger acceleration, faster change. C: Object 1 changes velocity slower: true. So pick C. D: Object 1's a is larger, Wrong: Object 2's is larger. The answer is C. Note: when comparing acceleration magnitudes, never let the sign sway you. Just look at the numbers. Next, if acceleration itself increases or decreases, what's the effect?

Only the direction of a matters, not its magnitude

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Note: to tell speeding up or slowing, we only check a's direction, not changes in its magnitude. Take this diagram: if v₀ and a point the same way, whether a grows or shrinks, it only changes the length of Δv. But whatever Δv's length, it's in v₀'s direction, so vₜ is bigger than v₀: speeding up. Likewise, if v₀ and a are opposite, a larger or smaller only changes Δv's length. Δv is opposite to v₀, so vₜ is shorter than v₀: slowing down. So whether acceleration grows or shrinks doesn't affect our judgment. If a and v point the same way, the object is speeding up, even if a is getting smaller, its velocity still increases, just more and more slowly.

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An example. This is t = 1 s, t = 2 s, t = 3 s, t = 4 s, t = 5 s. A car speeds up to the right, but acceleration decreases. Write down its velocity: at t = 1 s, velocity is 10 m/s; at t = 2 s, it reaches 15 m/s. at t = 3 s, maybe 17 m/s; t = 4 s, maybe 18 m/s; t = 5 s, 18.3 m/s. I made these up. But you can see: from t = 1 to 2 s, it rose by 5, from 2 to 3, by 2, from 3 to 4, by 1, from 4 to 5, only 0.3. So when acceleration is decreasing, from t = 1 s to 5 s, velocity still rises, just more and more slowly: each second adds less than the one before. When a = 0, it ends up in uniform motion.

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If a is opposite to v, even as a grows, velocity still decreases, but if a grows further, it falls faster and faster. Remember the key point: to tell speeding up or slowing, look only at direction, a's direction. Ignore whether acceleration's magnitude grows or shrinks; it doesn't affect that judgment at all. Direction alone is enough. In problems from now on, if you see acceleration rising or falling, don't let it mislead your judgment; Just ignore it. A few more to practice this.

Practice: house prices and a shrinking a

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Xiao Gang reads in the news that after China brought in housing-price controls, for a while, home prices in some cities kept rising, but more slowly. Xiao Gang: rising prices = speeding up, falling prices = slowing down. Pretty good! Learning physics from financial news. Based on this analogy, If house price growth is slowing down, what is that like in motion? Prices rising means speeding up, since rising is speeding up. So v is increasing at first. But slowing growth means it rises more and more slowly, like speed increasing more slowly. So a and v point the same way, but a is decreasing. So just pick B? v increases, since it's speeding up, but a decreases. So B.

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This one's a good problem. Many subjects are the same at heart, so analogies deepen your physics. It's a great method, one I love. I love taking a physics idea and finding a real-life example, like how quantities relate: Dad, grandpa, son. Then it all clicks. Easy one, nothing more to say. Next: which of these v and a relationships can exist? Let's go one by one. A: v points east and is decreasing. Decreasing means slowing down, so a and v are opposite. v points east, so a must point west. That's fine. Whether a is growing or shrinking is unknown. It could grow, or shrink; we can't tell which. So a growing or shrinking both work. A is correct.

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B: v points east and is decreasing, so it's slowing down, a and v opposite. v is east, so a is west. a can't point east, right? B is wrong. C: v east, and increasing. Increasing: a and v same direction. v is east, so a must be east, not west. C is wrong. D: east, and increasing, means a and v match. Speeding up, so a and v match. v is east, so a is east, not west. So D is wrong. The answer is A. If you haven't summed up a and v with "same speeds up, opposite slows," this one gets confusing. But with that idea, it's easy. Judging speeding up or slowing down is a very, very important skill. Practice it until you've nailed it. Answer: A. Not hard, so moving on. Now three core statements. These three are very handy for acceleration concept questions.

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You'll use them in your exams. They're tested a lot. Don't memorize these three. Just understand each one. Not hard. Simple.

Three core statements: a = 0, a changing, a's direction

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First: if an object has a = 0, it isn't necessarily at rest. a = 0 means v is unchanged, and that has two cases. Case 1: it's at rest, v is 0 all along. v constant: it's at rest. Case 2: uniform straight-line motion, which is motion with constant velocity. Velocity stays constant. So a = 0 doesn't mean at rest. If an option says "a = 0 means at rest," can we pick it? No, we can't. Second: if you only know whether a is increasing or decreasing, can you tell if v increases or decreases? You can't. Say a is increasing. Is v rising or falling? Can't tell. To tell if v rises or falls, check the direction of a and v. I only know how a changes, not whether a and v point the same way.

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If a grows and v is in a's direction, it speeds up. If a grows but v is opposite to a, it slows down. Note: if you only know a's trend, you can't tell v's trend. Third: if you only know a's direction, you can't tell v's direction. Say an object's a points right. Is v left or right? It could be left or right. If v is right, it speeds up to the right, if v is left, it slows down. So knowing only a's direction, you can't tell v's direction. See how much these three matter in problems.

School exam problems on a and v

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These problems are from school exams. This one is from Hengshui High School. On velocity and acceleration, which is correct? A: a is zero, so v must be zero. Isn't this the first of those three? a = 0: v = 0, or uniform straight-line motion So A is wrong. Now B and C: a is known to decrease. Can we tell if v rises or falls? That's my second statement: knowing only a's trend, can't tell v's. So neither B nor C. D: the larger a is, the faster v changes. Yes, a is the rate of change of velocity. So larger a, faster change: answer D. A, B and C all used what I just taught. Get a feel for it. Next, from my alma mater, Yinchuan No. 1 High. Which is false?

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A: a points east, v points west. Yes, possible. If a is east, v could be east or west. A is true, so we skip it. B: a = 0, so v must be 0. Wrong. That's statement one: a = 0 can be uniform motion, v ≠ 0. So B is false: we pick it. C: Object 1 has a bigger change in v, but a smaller a. Possible! The two times may differ. Example: objects A and B. A speeds up from 0 to 10 m/s, B from 0 to 5 m/s. A's change in velocity is bigger. But say A took 1000 s and B only 1 s. Who has the larger a? B's a is larger, clearly. So C is true. Bigger Δv doesn't mean bigger a; it also depends on time.

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D: motion in a straight line, a in the later stage is smaller, but v is larger. Possible? Yes: speeding up with a decreasing, a, v same way, But a is decreasing. So acceleration is getting smaller, so the later stage has less acceleration But a and v point the same way, so it speeds up and v keeps increasing It can be bigger than before. So D is possible too. So the answer is B. Next: Tianjin Yaohua High School. A: Zero acceleration means at rest. Not quite. At rest or constant velocity. So A is wrong. B: acceleration decreases gradually, velocity may keep increasing. If acceleration drops, v can rise or fall So B is correct. C: acceleration increases, so v must increase. Note "must": that's already wrong. I told you: if acceleration increases, v can go up or down.

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a and v may be same or opposite. So C is wrong. D: acceleration is 0, v may be very large. Like a plane flying in the sky, at constant velocity, v is large but acceleration is 0. That fits the description. The first half is right. Claim: when v = 0, a can't be large. Not true, right? Remember last lesson's rocket launch? The thrusters are firing downward, about to launch. Not moving yet. At this instant, v is 0. But the next instant, it has speed, v₁ might be 1000 m/s. Is acceleration 0? No. It's large too. So the second half is wrong. Note: v = 0 doesn't mean a = 0. So the answer is B.

Does displacement keep growing? Check for a turn-around

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Next, two more questions mixing v, a, and displacement. Common type. Like this: From the 2019-20 100-school alliance exam Sounds pretty cool. A particle moves in a straight line with unchanging velocity direction. It keeps going one way, say to the right, v direction never changes. Let me analyze this. Look at options A, B, and C. First two sentences are the same. If a and v are in the same direction, it's speeding up, since a and v align. Then as a decreases gradually to 0, it's speeding up with decreasing a. Still speeding up, but slower and slower, then constant velocity. When its acceleration decreases to 0, Is v at its minimum? No. It keeps speeding up all along, v keeps increasing. When a reaches 0, velocity stays constant.

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So v reaches its maximum, not its minimum. v first rises, then stays constant. That value is the maximum. So between A and B, pick B? v is at its maximum. C is about displacement. How do we tell if displacement grows? Magnitude of displacement is the distance from start to your current position. The farther from the start, the bigger the displacement. Never mind speeding up or slowing down. Here, v never turns: I just keep going. Going forward, aren't you moving away from the start all the time? So displacement keeps growing. At a = 0, I move at constant velocity. I still move farther away, so displacement still grows. So C is wrong. D: if a and v are opposite, it's slowing down. Then a decreases to 0. Is displacement minimal?

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Surely not. This is slowing down with decreasing a. I slow down first, then constant velocity at a = 0. Slowing or constant, I keep going forward. Just slower and slower. As long as I move on, displacement grows. Option D: displacement also keeps rising. Surely not the minimum. So D is wrong too. So only B is correct. How to tell if displacement increases? Note: magnitude of displacement is just your current distance from the start. If it always goes one way, displacement keeps increasing. But if it goes a while, then turns around, If it heads back, its distance from the start shrinks, So displacement decreases. Note: if it doesn't turn around, displacement keeps increasing It does not depend on speeding or slowing.

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Got displacement analysis now? So the answer is B. Next: Hebei Hengshui High School. A particle moves along a straight line. At t = t₀: s > 0, v > 0, a > 0. This wording isn't rigorous, As we know, for quantities like v and a, the sign doesn't mean above or below 0. It only shows: along or against positive. They mean a, v, and s are all positive. "Greater than 0" isn't rigorous, Just keep that in mind. So a and v point the same way, since both are positive. Then a decreases gradually to 0. At first a and v align, so it speeds up. Then a decreases to 0, still speeding up, but ever slower. It speeds up first, once a is 0, velocity stays constant.

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Overall: speed up, then constant. Look at the options. A: v changes more and more slowly. Isn't v's rate of change acceleration? Smaller a, v changes more slowly. Yes. At a = 0, v stops changing, Constant velocity. So A is correct. B: v decreases gradually. Wrong: v rises, then stays constant. I speed up, then go constant. v rises when speeding up; constant after. v never decreases, so B is wrong. C: displacement keeps increasing. I move one way, never turning around? So displacement keeps growing. So C is correct. D: displacement and v stay positive. Yes, I'm going the positive way, no turning around. so both are positive. So D is correct too. The answer is A, C, and D.

Opposite a and v: slow to zero, then reverse

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Now a special kind of motion. Tested often. Pay attention. Here's an example. Say the initial velocity points right, and the acceleration points left, and it never goes away. What motion will this object do? Its motion has two stages. Stage one: velocity points right, a points left, a and v are opposite, so it slows down. Since a never goes away, it keeps slowing down until v = 0. When it slows to 0, a is still there. Now this happens: It has stopped, v = 0, but a still points left. How does it move now? It turns around and moves left, speeding up again. Stage two: speeds up in reverse from 0. Now v and a both point left, right? a, v same way: it keeps speeding up.

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Overall, the motion has two stages. Stage one: slowing down, stage two: speeding up in reverse. This is another common kind of motion, so here's one more summary. More practice now. From Zunyi No. 4 High School, Guizhou. Which statement is correct? A: acceleration keeps decreasing, so does velocity. Wrong, right? We covered this. Knowing whether a rises or falls doesn't tell you velocity's trend. So A is wrong. B: bigger acceleration means bigger change in velocity? Wrong: it means velocity changes faster. Acceleration is how fast velocity changes. So B is wrong. C: if a's direction is fixed, so is velocity's direction. Careful: C is wrong. That's the scenario I just analyzed. Here my a stays left: constant. But velocity started rightward, then slowed to zero, then sped up leftward.

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Direction changed! I analyzed this scenario just for option C. If you've never summed up this scenario, you may have no clue on C. C is wrong. D: acceleration measures how fast velocity changes. No problem there. Pick D for this one. Let's really sum up this option C.

Exam problem: a shrinks to zero

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Next is from Hengshui High School, A moving object: if acceleration falls to 0, what could its velocity do? Several cases; which are possible? A: speed keeps decreasing. Possible! If a and v are opposite, it slows down, right? So A is possible. B: speed increases. Possible: a and v same way, even if a shrinks, it speeds up. So B is possible. C: speed decreases first, then increases. That's possible too. It's the process I covered earlier. Let's check with numbers. Say that at first, my velocity is 5 m/s to the right, and my acceleration is 10 m/s² leftward. So I'm slowing down. And as stated, my a keeps shrinking. Could the velocity hit 0 but a hasn't hit 0 yet?

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Say velocity is 0 now, and acceleration has dropped to 3 m/s². Possible? Yes. You never said who hits 0 first. Then what does it do next? Does it start speeding up leftward? Yes, since a points left. At v = 0 it's at rest, so it follows a's direction, speeding up to the left. So speed decreases, then increases? So C is possible too. Pick C. D: increases, then decreases. Impossible. If it speeds up first, then a and v are the same way. And when a falls to 0, it moves uniformly, never slowing. Once a is 0, there's no acceleration, so it speeds up, then goes steady. So D is impossible. A, B, C are all possible. This one is a bit tougher. To end this lesson,

Uniform acceleration on a line: three cases

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let's define a very important motion: Constant-acceleration linear motion, high school's most important motion. 90% of calculation problems use it. What is that motion? Linear motion, constant acceleration. Constant acceleration means its magnitude and direction both stay. That's constant-acceleration motion. Its velocity changes uniformly, because acceleration describes how fast velocity changes. If a stays constant, velocity changes uniformly. It has three possible cases. First: a and v same way, speeding up. We call it uniform speeding up. Second: a and v opposite, slows to 0 and the motion ends. That's uniformly decelerated motion. Third: it slows down to 0, but acceleration is still there, so it then speeds up the other way.

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First uniform slowing, then speeding up. It's the two-stage process from before. Constant-acceleration motion is an umbrella term for both kinds. So uniform speeding up and uniform slowing down both fall under it. Now, the last problem of this lesson. This lesson has quite a few problems, but I can't bear to cut any, I want you to practice each. So hang in there.

Last practice: statements about a and v

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From Hengshui High School, Hebei. In straight-line motion, which statement about a and v is correct? A: if a's direction is constant, velocity's direction may change. Possible, like the last example: a and v opposite, two stages, first slows, then speeds up backward, so v's direction flips, So A is possible. B: in uniform deceleration, a must be negative. Careful: not necessarily, it depends on your positive direction. Say an object's velocity points rightward, acceleration leftward, but you choose leftward as positive, then a is positive! So B is wrong. Honestly, a dull option. C is wrong too; we covered it before. Remember the rocket? Velocity is zero for now, but a is huge, so velocity soon grows. So when velocity is zero, a isn't necessarily zero.

48:17
D: only if the initial velocity direction is positive can acceleration be positive when it's uniformly speeding up. Note: if it says it's uniformly speeding up, then a, v are same way. If initial velocity is positive, a is positive; if it's negative, a is negative. To get positive a, v₀ must be positive. So D is correct too. What D really tests: do you know, in speeding up, a and v same way. That's the point. That's it for this problem.

Summary

48:56
Finally, the Summary. First, the relation of a, v, and Δv. Though it may not help much with directly solving problems, it matters a lot for understanding and really learning physics. It's very important. Later we'll often wonder: how do the quantities relate? Remember what I taught: Find cause and effect: who affects whom. Who's dad? Who's son? Important! Second, this lesson's most important point: speeding up vs. slowing down. Core: check a and v directions together. Same way: speeds up. Opposite: slows. Third: our three core sentences, great for solving problems. First: when a is 0, it isn't necessarily at rest; it may move uniformly. Second: knowing only how a changes can't tell velocity's trend. Third: a's direction alone can't tell v's. Last: constant-acceleration linear motion.

49:59
The definition is easy: linear motion, constant a. Key: three possible cases. Speed up all along, or slow to 0 and stop, or slow to 0, then speed up backward. The third is easy to overlook. That's all on acceleration for now. It's the star of high school physics. It will show up in countless lessons. We'll stop here for this lesson.