1.1 Displacement, Velocity and Acceleration — Teaching Script

AP Physics C: Mechanics · Unit 1 Kinematics

This is a spoken lecture script designed for practice delivery. Stage directions in brackets. Speak naturally; formulas in $$ are what you write on the board as you talk.

Opening

Alright, everyone. Today we're starting Unit 1 — Kinematics. And I want you to remember one thing before we even begin:

Kinematics is just a language. It's the language we use to describe motion. We're not explaining why things move yet — that's dynamics, that's Newton's laws, that's next unit. Today, we're just learning how to talk about motion precisely.

So if you ever feel like this unit is just a bunch of definitions and formulas and graphs... you're right. But here's the thing — every single thing we do for the rest of the year depends on this language. So let's get it right.

[pause]

Part 1: Position, Displacement, and Distance

Position

Let's start with the simplest question: where is the object?

We pick a coordinate system — an origin, a positive direction — and we write down the object's coordinate, $x(t)$. That's it. That's position.

[write on board: $x(t)$]

Now, here's something that seems obvious but trips people up: the number you get for position depends on where you put the origin. If I stand here and call this zero, the object is at plus three. If I move my origin over there, suddenly it's at minus two. Position is relative. The object didn't move — you moved your reference frame.

[pause]

Displacement vs. Distance

OK, now the object moves from point A to point B. How do we describe that?

Well, there are two ways, and they give you different answers:

[write on board: $\Delta x = x_2 - x_1$]

This — displacement — is just "final position minus initial position." It's a vector. It has a sign. And it does not care about the path. Straight line, zigzag, loop-de-loop — displacement only looks at where you started and where you ended.

Distance, on the other hand — that's the total length of the path you actually traveled. It's a scalar. It's always positive. And it absolutely cares about the path.

[pause]

Let me give you the classic example. You run one full lap around a 400-meter track. You start here, you run all the way around, you come back to the same spot.

What's your displacement?

[wait for answer]

Zero. You started here, you ended here — same position, so $\Delta x = 0$.

What's your distance?

400 meters. Your feet actually moved 400 meters.

[write on board: $\Delta x = 0$, $d = 400\ \mathrm{m}$]

This is the #1 way AP multiple choice tries to trick you. They'll say "a person walks 3 meters east, then 4 meters west" — and they'll ask for both displacement and distance. Displacement is negative 1 meter... or, well, depends on which direction you call positive. Distance is 7 meters. Don't mix them up.

[pause]

Part 2: Velocity

Average Velocity

Now let's add time to the picture. The object moved from here to there — how fast did it do that?

[write on board: $\bar{v} = \frac{\Delta x}{\Delta t}$]

Average velocity is displacement over time. Notice — it's displacement in the numerator, not distance. So if you did a round trip and your displacement was zero, your average velocity is also zero. You went somewhere and came back — on average, you didn't go anywhere.

[pause]

Average Speed

Average speed is the other one:

[write on board: $\text{average speed} = \frac{d}{\Delta t}$]

Distance over time. And here's the key point — unless the object is always moving in the same direction, average speed is not equal to the absolute value of average velocity. They look similar, they sound similar, but they're different quantities with different numerators.

Round trip: average velocity is zero, average speed is definitely not zero. That's the difference.

[pause]

Instantaneous Velocity

Now, average velocity is nice, but what if I want to know how fast you're going right now, at this exact instant?

Well, I take a small time interval, $\Delta t$, and I shrink it... and shrink it... and shrink it... until it approaches zero. In the limit, I get the instantaneous velocity:

[write on board: $v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$]

If you've taken calculus, you recognize this — that's a derivative. $v = \frac{dx}{dt}$.

Now, what does this mean geometrically? If I plot position $x$ against time $t$:

[pause — gesture as if pointing to a graph on the board]

The derivative at any point is the slope of the tangent line at that point. Positive slope — you're moving in the positive direction. Negative slope — you're going the other way. Zero slope — you're momentarily at rest.

But — and this is important — zero velocity does not mean zero acceleration. Think about a ball thrown straight up. At the very top, for one instant, $v = 0$. But gravity is still pulling down. The acceleration is still $-g$. We'll come back to this.

[pause]

Part 3: Acceleration

Definition

So velocity can change. How fast does it change? That's acceleration.

[write on board: $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$]

Acceleration is the derivative of velocity — or the second derivative of position. It's the rate of change of velocity.

Now listen carefully: acceleration is not "how fast you're going." That's speed. Acceleration is "how fast your speed is changing." You can be going incredibly fast with zero acceleration — cruising at 200 km/h on a highway, constant velocity, $a = 0$. You can also be going very slowly but accelerating hard — a car starting from a red light, $v$ is small but $a$ is large.

[pause]

How to Read the Sign

OK, this is where most students lose points. Listen carefully.

If I say "the acceleration is negative" — does that mean the object is slowing down?

[wait — let them think]

Not necessarily.

The rule is: you have to look at the relationship between $a$ and $v$. Not $a$ alone.

[write on board:]

$a$ and $v$ same sign → speeding up
$a$ and $v$ opposite sign → slowing down
$a = 0$ → constant velocity

Here's the counterexample that breaks everyone's brain:

[write on board: $v = -8\ \mathrm{m/s}$, $a = -2\ \mathrm{m/s^2}$]

Both are negative. The object is moving in the negative direction, and it's getting faster. It's accelerating — speeding up — in the negative direction. The acceleration is negative, but the object is not decelerating.

So burn this into your brain: negative does not mean slowing down. Same sign means speeding up. Opposite sign means slowing down. That's it.

[pause]

Common Misconceptions

Let me rattle off three things I hear every year that are wrong:

One — "negative acceleration means deceleration." Wrong. You need to check the sign of $v$ first.

Two — "if velocity is zero, acceleration must be zero." Wrong. Ball at the top of its flight, $v = 0$ but $a = -g$. Gravity doesn't take a coffee break.

Three — "if acceleration is decreasing, the object must be slowing down." Wrong again. If $a$ is decreasing but still has the same sign as $v$, the object is still speeding up — just not as quickly as before. It's like easing off the gas pedal — you're still accelerating, just less aggressively.

[pause]

Part 4: The Three Graphs

Now, we have three graphs: $x$ vs $t$, $v$ vs $t$, and $a$ vs $t$. And the relationship between them comes down to one sentence:

[write on board: Slope = derivative. Area = integral.]

That's it. That's the whole section.

[draw three graph sketches side by side on the board]

On the $x\text{-}t$ graph: the slope gives you velocity. The area doesn't mean anything useful.

On the $v\text{-}t$ graph: the slope gives you acceleration. The area under the curve gives you displacement — and here's the catch: area below the axis is negative. That's the most common mistake on graph problems — students see a triangle below the axis and add it instead of subtracting.

On the $a\text{-}t$ graph: the slope is the jerk — rate of change of acceleration, not on the AP exam very often but good to know. The area gives you the change in velocity, $\Delta v$.

[pause]

So in equation form:

[write on board: $v(t) - v(0) = \int_0^t a\,dt \qquad x(t) - x(0) = \int_0^t v\,dt$]

The area under $a\text{-}t$ gives $\Delta v$. The area under $v\text{-}t$ gives $\Delta x$. Slope of $v\text{-}t$ gives $a$. Slope of $x\text{-}t$ gives $v$. Slope = up one level, area = down one level. Memorize this pattern.

[pause]

Part 5: The Four Kinematic Equations

Now, when acceleration is constant — and only when it's constant — we can integrate $a = \frac{dv}{dt}$ twice and get four equations:

[write on board, one at a time:]

$$v = v_0 + at$$

$$\Delta x = v_0 t + \tfrac{1}{2}at^2$$

$$v^2 = v_0^2 + 2a\Delta x$$

$$\Delta x = \frac{v_0 + v}{2}\,t$$

Now, before I tell you how to pick which one to use, let me warn you: these only work when $a$ is constant. If the problem says "the acceleration is proportional to velocity" — like $a = -kv$ for air resistance — these four equations are garbage. Useless. You have to go back to calculus and integrate.

AP Physics C will test this. They'll give you a variable acceleration just to see if you blindly plug into these formulas or if you actually think.

[pause]

OK, how to pick:

The problem doesn't mention time $t$? Use $v^2 = v_0^2 + 2a\Delta x$.
The problem doesn't give you final velocity $v$? Use $\Delta x = v_0 t + \frac{1}{2}at^2$.
The problem doesn't give you acceleration $a$? Use $\Delta x = \frac{v_0 + v}{2}\,t$.

Each equation is missing one variable. Match the missing variable to the equation. Simple.

[pause]

Part 6: Free Fall

Free fall is just a special case of constant acceleration. We throw away air resistance, and then everything — a feather, a bowling ball, a rocket — falls with the same acceleration:

[write on board: $a = -g \approx -9.8\ \mathrm{m/s^2}$]

Now, the sign depends on your coordinate system. If I call up positive, then $a = -9.8$. If I call down positive, then $a = +9.8$. Either way is fine — but pick one and stick with it for the entire problem. Don't switch halfway through, that's chaos.

[pause]

Here's the scenario that always confuses students: you throw a ball straight up at, say, $20\ \mathrm{m/s}$.

Going up: velocity is positive, acceleration is negative — opposite signs — the ball is slowing down. Good.

At the top: velocity is zero. But acceleration? Still $-g$. Still $-9.8$. Gravity doesn't turn off at the top. The ball is momentarily at rest but still accelerating downward.

Coming down: velocity is now negative — same sign as acceleration — the ball is speeding up. In the negative direction.

And here's a beautiful symmetry: the time going up equals the time coming down. If it takes 2 seconds to reach the top, it takes 2 seconds to come back. And the speed when it returns to your hand is the same as the launch speed — just directed downward.

[pause]

Part 7: Worked Examples

Example 1: When Displacement ≠ Distance

[write on board: $x(t) = 2t^3 - 6t^2 + 4$ meters]

A particle moves along the $x$-axis. Its position is $x(t) = 2t^3 - 6t^2 + 4$ meters. I want three things: the velocity function, the acceleration function, and — from $t = 0$ to $t = 3$ seconds — both the displacement and the distance.

OK, part (a). Velocity is the derivative:

[write on board: $v = 6t^2 - 12t = 6t(t-2)$, $a = 12t - 12$]

Part (b), displacement. Just plug in the endpoints:

$x(0) = 4$. $x(3) = 2(27) - 6(9) + 4 = 4$. So $\Delta x = 4 - 4 = 0$.

Displacement is zero. The particle ends up where it started.

[pause]

But part (c), distance — that's the trap. Because if displacement is zero, you might think "well, it didn't move." Wrong. Let's check where the velocity is zero — $v = 6t(t-2) = 0$ at $t = 0$ and $t = 2$. So the particle reverses direction at $t = 2$.

At $t = 2$: $x(2) = 2(8) - 6(4) + 4 = -4$.

So the particle goes from position 4, down to $-4$, then back up to 4. That's $8$ meters down plus $8$ meters back up:

[write on board: $d = |{-4 - 4}| + |{4 - (-4)}| = 8 + 8 = 16\ \mathrm{m}$]

Distance is 16 meters. Displacement is zero. Same motion, two different answers, depending on what you're asking.

[pause]

Example 2: Braking

A car is going $30\ \mathrm{m/s}$. The driver brakes, and it takes $5$ seconds to come to a complete stop. Find the acceleration and the braking distance.

This is straightforward constant-acceleration. Take the direction of motion as positive.

[write on board: $a = \frac{0 - 30}{5} = -6\ \mathrm{m/s^2}$, $\Delta x = \frac{30+0}{2}\times 5 = 75\ \mathrm{m}$]

$a = -6\ \mathrm{m/s^2}$. Negative — because it's opposing the motion. The car is slowing down. Opposite signs, confirmed.

Distance: $75$ meters. That's why tailgating is a bad idea.

[pause — light laugh]

Example 3: Vertical Toss

A ball is thrown straight up at $20\ \mathrm{m/s}$. Find the maximum height and the total time to return to the launch point. Use $g = 9.8$.

Up is positive, so $a = -9.8$. At the highest point, $v = 0$.

[write on board: $0 = v_0^2 - 2gh \Rightarrow h = \frac{400}{19.6} \approx 20.4\ \mathrm{m}$]

Maximum height: about $20.4$ meters.

For the time: when the ball returns to the launch point, displacement is zero. So:

[write on board: $0 = v_0 t - \frac{1}{2}gt^2 \Rightarrow t = \frac{2v_0}{g} = \frac{40}{9.8} \approx 4.1\ \mathrm{s}$]

About $4.1$ seconds total — roughly $2$ seconds up, $2$ seconds down.

[pause]

Part 8: Common AP Mistakes — Rapid Fire

Let me rattle off the five mistakes that cost the most points:

Mixing up average velocity and average speed. Round trips — the difference is huge.
Forgetting negative area on a $v\text{-}t$ graph. Below the axis means negative.
Thinking $a < 0$ means slowing down. Check the sign of $v$ first. Always.
Using the kinematic equations when $a$ isn't constant. If $a$ varies, you integrate.
Sloppy with signs and units. Pick your positive direction, write it down, stick with it.

[pause]

Closing

So that's kinematics in a nutshell. Let me summarize:

Displacement is a vector — start to finish. Distance is a scalar — total path.
$v = \frac{dx}{dt}$, $a = \frac{dv}{dt}$. Slope on a graph.
Area under $v\text{-}t$ is displacement. Area under $a\text{-}t$ is change in velocity.
Four kinematic equations — constant acceleration only.
Speeding up or slowing down? Same sign or opposite sign. That's the question.

Next time, we'll take this into two dimensions. Same ideas, more vectors. Read through your notes tonight, and try the homework problems.

Homework

A particle moves along a straight line with $v(t) = 3t^2 - 12t + 9$ ($\mathrm{m/s}$). Find the displacement and distance from $t = 0$ to $4\ \mathrm{s}$.
A ball is dropped from a height of $45\ \mathrm{m}$. Find the time to reach the ground and the impact velocity ($g = 9.8\ \mathrm{m/s^2}$).
Given $a(t) = 6t$ ($\mathrm{m/s^2}$), initial velocity $v(0) = 2\ \mathrm{m/s}$, initial position $x(0) = 0$. Find $v(t)$ and $x(t)$, and explain why the constant-acceleration equations cannot be used directly.