Kinematics

LANGUAGE FIRST: THE FOUR QUANTITIES

Physics uses several ordinary English words in a narrower sense than you do, and most early lost points are vocabulary, not physics.

Position $x$ is where you are, measured from an origin you chose. It is not “how far you’ve gone.” Position can be negative; that just means you are on the other side of the origin from the direction you called positive.

Displacement $\mathrm{\Delta }x=x_f-x_0$ is the change in position, and it carries a sign. Distance is the odometer reading: total path length, never negative, never subtracting. Distance equals the size of the displacement as long as you never turn around, and the moment you do turn around they part ways permanently.

Walk 6 m forward, then 4 m back. Distance is $6+4=10$ m — your feet did all of it. Displacement is $+2$ m — you finished 2 m from where you began, and the 4 m back canceled 4 m of the outbound trip. Displacement doesn’t care what route you took or how many times you doubled back; it compares two points and nothing else.

Average velocity $={\displaystyle \frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}}$ uses displacement, so it inherits the sign and the not-caring-about-route. Average speed $={\displaystyle \frac{\text{distance}}{\mathrm{\Delta }t}}$ uses distance, so it is never negative. For the walk above over $9$ s: average velocity $=2∕9=+0.22$ m/s; average speed $=10∕9=1.1$ m/s. Same trip, same clock, two answers differing by a factor of five. When a question says “velocity,” it wants the signed one, and the sign is part of the answer.

Instantaneous velocity $v$ is the velocity right now rather than averaged over an interval — the number a speedometer shows, with a direction attached. On an $x$–$t$ graph it is the slope at that instant.

Acceleration $a={\displaystyle \frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}}$ is how fast the velocity changes. Across an interval that ratio is the average acceleration; the instantaneous acceleration is the slope of the $v$–$t$ graph at a single instant — the same distinction velocity just drew. This unit assumes the acceleration is constant, which is the one case where the two are the same number. A $v$–$t$ graph that curves is telling you they aren’t. Read the definition again, because the trap is in it: acceleration is not speed, and it is not “speeding up.” A car braking from $25$ m/s to rest in three seconds has a large acceleration. A car cruising at $60$ m/s has none. Acceleration measures change, so a fast object that isn’t changing has zero acceleration and a slow object that’s changing quickly has a lot. An object slowing down has an acceleration pointed against its motion — which is why acceleration needs a sign of its own, independent of the sign of the velocity, and why the two signs together are worth their own toolbox entry below.

READING MOTION GRAPHS — THE HIGHEST-VALUE SKILL IN THE UNIT

Three graphs describe the same motion. They are not three unrelated pictures; they are one story told at three levels, and exactly two operations move you between them.

One motion, three graphs, aligned on the same clock. Read downward with slopes, upward with areas.

Going down (slope). The slope of the $x$–$t$ graph at any instant is the velocity at that instant. That isn’t a rule to memorize — it is the definition, arriving in graph form. Slope means rise over run; the rise on an $x$–$t$ graph is $\mathrm{\Delta }x$ and the run is $\mathrm{\Delta }t$, so slope is $\mathrm{\Delta }x∕\mathrm{\Delta }t$, which is precisely what velocity means. Check the units and you’ll never doubt it again: meters on the vertical divided by seconds on the horizontal gives m/s. The same argument one level down: the slope of the $v$–$t$ graph is $\mathrm{\Delta }v∕\mathrm{\Delta }t$, in (m/s)/s $=$ m/s${}^2$, which is acceleration.

Going up (area). The area under the $v$–$t$ graph is the displacement. Here is why, and it is worth seeing once rather than accepting. Suppose the velocity were constant at $v$ for a time $\mathrm{\Delta }t$. Then $\mathrm{\Delta }x=v\mathrm{\Delta }t$ — and on the graph, $v$ is the height and $\mathrm{\Delta }t$ is the width, so $v\mathrm{\Delta }t$ is literally the area of that rectangle. Now take a velocity that is changing: chop the time axis into slices thin enough that the velocity barely changes across any one of them. Each slice contributes its own little rectangle of displacement, and the total displacement is the sum of all of them — which is the area under the graph. Units again: (m/s) $\times$ s $=$ m. Height times width gives meters, so the area under a $v$–$t$ graph could not be anything but a displacement. One level down, identically: the area under the $a$–$t$ graph is $\mathrm{\Delta }v$, the change in velocity (m/s${}^2$ $\times$ s $=$ m/s).

Now read the figure across the dashed lines, which is the part worth your time. In the first phase $a$ is a positive constant, so $v$ is a straight line climbing at a constant rate, so $x$ curves upward — the slope of $x$ is itself getting bigger. In the middle phase $a=0$, so $v$ is flat, so $x$ is straight: unchanging slope, unchanging velocity. In the last phase $a$ is negative, $v$ falls back toward zero, and $x$ levels off.

Notice what the last phase is not. The object does not move backwards there. Its velocity stays positive right up to the final instant — look at the $v$ panel, which is above the axis the whole time — so $x$ keeps increasing; it just increases more and more slowly. Negative acceleration with positive velocity means slowing while still going forward. That one sentence resolves a large fraction of all sign confusion in this unit, and the figure is where you can see it rather than take it on faith.

SIGNED AREA, AND THREE HABITS THAT PREVENT MOST GRAPH ERRORS

Area below the time axis counts as negative displacement, because the object is moving backwards during that stretch and giving back ground it had already gained.

Displacement $=+16-4=+12$ m. Distance traveled $=16+4=20$ m. Add the areas’ sizes for distance; add them with signs for displacement.

That picture is the whole distance/displacement distinction in one place. If a question asks for displacement, subtract the area below the axis; if it asks for distance traveled, add it. And note where the turnaround happens — not where the graph is steepest, not where the graph is highest, but where it crosses zero. A zero crossing on a $v$–$t$ graph is the only place an object reverses.

Three habits:

• Check which graph you’re on before you interpret a feature. A flat line means “not moving” on $x$–$t$ and “moving at a steady speed” on $v$–$t$ — opposite situations, identical appearance. The most expensive graph error there is reads a $v$–$t$ graph as though it were $x$–$t$: a line rising from the origin looks like “moving away steadily,” but on $v$–$t$ it means “speeding up from rest.”
• A value and a slope answer different questions. “Where is it moving fastest?” on an $x$–$t$ graph means “where is it steepest,” not “where is it highest.” The highest point of an $x$–$t$ graph is usually a place where the object has stopped.
• Below the axis is not “less” — it is “the other way.” A velocity of $-4$ m/s is the same speed as $+4$ m/s, pointed opposite. Rank speeds by distance from the axis, not by height on the page.

THE BIG THREE, AND HOW TO CHOOSE AMONG THEM

These apply only when the acceleration is constant. Every one of them is a statement about five quantities: displacement $\mathrm{\Delta }x$, starting velocity $v_0$, final velocity $v$, acceleration $a$, and elapsed time $t$.

Plus a fourth that most courses under-use: $\mathrm{\Delta }x=\frac{1}{2}(v_0+v)t$. It is just “average velocity times time” — and when the acceleration is constant, the average velocity really is the plain midpoint of the starting and ending velocities, which is why the formula is so simple.

Here is the structural fact that makes choosing easy. There are five variables, and every equation involves exactly four of them. So each equation has a hole — one variable it never mentions. Line them up and the selection rule stops being something you memorize:

Filled dot: the equation contains that variable. Open ring: it doesn’t. Find the variable you neither know nor want — there is exactly one row without it.

Read the open rings down the diagram: $\mathrm{\Delta }x$, then $v$, then $t$, then $a$. Four equations, four different holes. Whichever variable you neither know nor want, exactly one row is missing it, and that row is your equation. You never have to solve simultaneously, and you never have to guess.

The procedure, every time:

1.

Write the five variables in a column and fill in every value you know, with its sign. Leave blanks where you don’t know.

2.

Circle the one the question asks for.

3.

Look at what’s left. One blank is the variable you neither know nor need. Go to the row without it.

Work an example against Problem 3’s braking phase. Known: $v_0=25$ m/s, $v=0$, $a=-7.5 {\text{m/s}}^2$. Wanted: $\mathrm{\Delta }x$. Left over: $t$ — you don’t know it and the question didn’t ask for it. So use the row with no $t$: $v^2=v_0^2+2a\mathrm{\Delta }x$. That took four seconds and no algebra, and it is the entire skill.

Two things people get wrong about these equations. First, constant acceleration is a condition, not a decoration. If the acceleration changes partway through — a car that cruises and then brakes — you cannot run one Big Three equation across the whole story. Break the motion into pieces, each with its own constant acceleration, and hand the ending velocity of one piece to the next as its $v_0$.

Second, $\mathrm{\Delta }x$ is displacement, not distance. If the object reverses inside the interval you’re using, the equation reports where it finished, not how far it traveled. When you need distance and a reversal happened, split the interval at the turning point (where $v=0$) and add the sizes.

SIGN CONVENTIONS: CHOOSE ONCE, THEN LIVE WITH IT

Kinematics in one dimension has no arrows, only signs — a sign is a direction. Which means the very first line of your solution should be a decision, written down: I am calling this direction positive. Up, right, east, downfield — whatever suits the problem. It does not matter which you pick. It matters enormously that you pick before you start and never revisit it.

Once chosen, everything else follows and nothing is negotiable. If up is positive, then a ball’s launch velocity is $+15$ m/s, gravity is $-9.8 {\text{m/s}}^2$, and a point 20 m below the launch is at $\mathrm{\Delta }y=-20$ m — including while the ball is on its way up, including at the peak, including on the way down. The acceleration does not flip when the motion does.

The confusion that costs the most points is treating the sign of $a$ as a synonym for “slowing down.” It isn’t. The sign of $a$ tells you which way the acceleration points; whether the object speeds up or slows down comes from comparing that direction to the direction of $v$:

Same signs $\Rightarrow$speeding up. Opposite signs $\Rightarrow$ slowing down. That is the whole rule, and it does not care which direction you called positive.

Two sentences worth stealing, because they are what a sign flip would wrongly claim:

• Writing $a=+9.8 {\text{m/s}}^2$ for a falling ball, after choosing up as positive, asserts that gravity pulls upward. Nothing in the problem said that, and nothing ever will.
• Writing $\mathrm{\Delta }y=+20$ m for a landing point below the launch asserts that the ball finished 20 m higher than it started. The ball fell into a river. Read your own equation back as an English sentence and errors like this announce themselves.

You may choose a different convention for a different problem — and Problem 6 deliberately takes down as positive, because everything in that problem moves downward and the arithmetic is cleaner with no minus signs at all. That is a fresh decision made at the start of a new problem, which is allowed. Changing your mind halfway through a problem is not.

FREE FALL — AND WHERE ITS SYMMETRY COMES FROM

“Free fall” means gravity is the only thing acting. In AP Physics 1 that covers essentially everything thrown, dropped, kicked, or launched, because the course tells you to neglect air resistance unless it says otherwise. The consequences:

• Choose up as positive; then $a=-g=-9.8 {\text{m/s}}^2$ for the entire flight — on the way up, at the top, on the way down.
• Free fall does not mean “falling.” A ball still rising is in free fall. The name describes what forces act, not which way the object happens to be going.
• At the peak, $v=0$ but $a$ is still $-g$. This is the single most-tested conceptual point in the unit, so be clear on why. Velocity passes through zero the way a thrown ball’s height passes through any other value — instantaneously, without stopping. If $a$ really were zero at the peak, the velocity would have no reason to change, and the ball would hang there forever. It doesn’t. What pauses at the top is the velocity’s value, not its rate of change.
• Mass is not in any of this. A heavy ball and a light ball thrown identically do identical things. If a problem gives you a mass in a kinematics question, it is either scenery or setup for a later part.

The symmetry facts, derived rather than asserted. Take a launch at $v_0$ upward from some height, and let the object return to that same height.

Time up equals time down. Going up, the velocity has to get from $v_0$ to $0$, and $v=v_0+\mathrm{at}$ gives $t_{\text{up}}=v_0∕g$. Coming back down from the peak, the velocity starts at $0$ and falls to some $v$ over the same drop $\mathrm{\Delta }y=-H$. But the trip up covered exactly that $H$ too, so the two legs are the same journey run backwards with the same constant acceleration. Formally: from the peak, $H=\frac{1}{2}g{t}_{\text{down}}^2$, and from the launch, $H=v_0^2∕2g$. Setting those equal, $t_{\text{down}}=\sqrt{2H∕g}=\sqrt{v_0^2∕g^2}=v_0∕g=t_{\text{up}}$.

Same speed at the same height. Use the equation with no $t$ in it, between launch and return, where $\mathrm{\Delta }y=0$ because you’re back where you started:

Same size, opposite sign: same speed, opposite direction. Notice the argument never used the specific numbers, so it holds at every height, not just the launch height — pick any $\mathrm{\Delta }y$ and the two crossings of that height give the same $v^2$.

Where the symmetry stops. It only holds between two points at equal height. A ball thrown from a bridge and landing in the water below has a symmetric rail-to-rail portion and a completely unsymmetric rail-to-water portion. Problem 4 uses symmetry for the first and honest algebra for the second; knowing which is which is the actual skill.

VECTOR COMPONENTS — THE TRIG THESE PROBLEMS NEED

Two-dimensional problems start by splitting one velocity into two, and end by putting two back together. That is all the trigonometry the unit requires, but you need it cold.

Splitting (launch). A velocity of magnitude $v_0$ at angle $𝜃$ above the horizontal has

Cosine goes with the horizontal and sine with the vertical when the angle is measured from the horizontal, which it almost always is. Rather than memorizing which is which, check with a limiting case, which takes two seconds and never lies: if $𝜃=0$ the launch is flat, so all the speed should be horizontal — and $\cos 0=1$, $\sin 0=0$ delivers exactly that. If the angle is instead given from the vertical, the roles swap, and the same limiting-case check catches it.

Recombining (impact). Given components $v_x$ and $v_y$, the speed is the hypotenuse and the direction is the arctangent:

Two sanity checks on the recombination. The speed must come out larger than either component — a hypotenuse always beats its legs. And the angle must land on the side you expect: if $\left|v_y\right|>\left|v_x\right|$ the motion is steeper than $45^{\circ }$, and if $\left|v_y\right|<\left|v_x\right|$ it is shallower. State the direction in words along with the number (“$60^{\circ }$ below the horizontal”), because a bare angle is ambiguous and loses credit.

Why components are legal at all. The horizontal and vertical directions are independent: what gravity does to $v_y$ has no effect whatsoever on $v_x$, and how fast the object travels sideways has no effect on how long it takes to fall. That independence is what lets you treat one two-dimensional problem as two one-dimensional problems, and it is the reason a marble rolled twice as fast off a table lands twice as far away in exactly the same amount of time.

PROJECTILES: THE TIME LIVES IN THE VERTICAL

Once decomposed, a projectile is two problems you already know, running side by side on one shared clock:

• Horizontal: $x=v_{0x}t$. Constant velocity. No acceleration, ever. $v_x$ at landing equals $v_x$ at launch.
• Vertical: free fall, with $v_{0y}$ as the starting velocity and $a=-g$ throughout.
• Shared: time, and only time. The two directions communicate through $t$ and through nothing else.

The consequence you use constantly: the time is in the $y$. Flight time is set entirely by the vertical story — launch height, launch vertical speed, gravity. The horizontal direction contributes nothing to it and simply spends whatever time the vertical hands over. So the standard solution order is fixed: decompose, solve the vertical for $t$, then use $t$ horizontally.

The other consequence: “at the wall” questions run backwards through the same channel. When a question asks about a specific place — “does it clear the fence at 55 m,” “is it above the net,” “where does it hit the wall” — you cannot answer from the vertical equation directly, because that equation speaks in time, not in $x$. The conversion is always the same two moves:

Position to time to the other position. Every “at the wall” problem in the course is that chain, and Problem 7 walks it.

Two useful landmarks. At the top of the arc, $v_y=0$ but $v_x$ is unchanged — so the projectile is still moving, horizontally, at $v_{0x}$. Its speed there is a minimum, and for any launch with $v_{0x}\ne 0$ it is never zero. (Straight up is the exception, and it is why Problem 4’s ball does stop at the top: with $v_{0x}=0$ there is nothing left to keep moving.) And for a launch and landing at the same height, the vertical symmetry facts carry over intact: the flight time is twice the time to the peak, and the landing speed equals the launch speed with $v_y$ reversed.

TWO OBJECTS, ONE CLOCK

When two things move and the question is about them meeting, passing, or catching, the entire method is bookkeeping:

1.

Choose one origin and one $t=0$, and use them for both objects. If one object starts 100 m ahead, that shows up as its $x_0=100$, not as a separate coordinate system.

2.

Write a full position function for each: $x_1\left(t\right)$ and $x_2\left(t\right)$.

3.

Translate the question into an equation about those functions.

Step 3 is where the thinking is, so be precise about the translation. “Catches up,” “meets,” “passes,” and “collides” all mean same position: $x_1=x_2$. “Same speed” means $v_1=v_2$, which is a different equation with a different answer, and the two moments are generally nowhere near each other. Confusing them is the classic error on this problem type, and Problem 5 draws the picture that makes the difference obvious.

Then interpret every root. Setting two position functions equal usually produces a quadratic, so you get two times. Neither is automatically garbage. A root of $t=0$ typically means “they were together at the start,” which is true and is not the answer you want. A negative root usually means “if this motion had been running before the problem began, they’d have met then” — physically meaningless here, so discard it, but say why. Two positive roots means they genuinely meet twice, and both are real answers. Never discard a root silently; name what it says first.

LINEARIZING DATA — TURNING A CONSTANT INTO A SLOPE

Lab questions hand you a table and want a physical constant. The move is always the same, and it has three steps.

1. Write the model. What relationship should hold if the physics is right? For a dropped object, $h=\frac{1}{2}g{t}^2$.

2. Choose axes that make the model a straight line. Compare the model to $y=\mathrm{mx}+b$. Here, if you plot $h$ on the vertical and $t^2$ on the horizontal, the model reads $h=\left(\frac{1}{2}g\right)\cdot t^2+0$ — a line through the origin with slope $\frac{1}{2}g$. You have to make a new column for $t^2$; that column is the work.

3. Read the constant out of the slope. Slope $=\frac{1}{2}g$, so $g=2\times \text{slope}$. Note that the slope, not any individual data point, is the measurement — that is the point of graphing at all.

Why bother, when you could just solve for $g$ from one row? Because a straight line is testable and a single row isn’t. If the points fall on a line, the model is behaving; if they curve away, your model is wrong or something systematic is interfering, and you can see it. And the slope uses all the data at once, so a single sloppy measurement moves it far less than it would move a one-row calculation. A curve can be neither verified nor measured reliably by eye; a line can be both.

The directional-error question, which is a separate skill. Lab questions reliably follow up with: if this measurement were off in this direction, would the result come out too high or too low? Answer it by tracing, not by intuition:

1.

Which measured quantity does the error affect, and which way?

2.

Where does that quantity sit in the formula — numerator or denominator?

3.

Therefore the computed result goes which way?

Then write the conclusion in that exact shape: “this makes the measured value too high because…” Graders are looking for the chain, not the verdict.

VERIFICATION HABITS — HOW TO CHECK WITHOUT REDOING

Every solution in this guide ends with at least one of these, and you should make them reflex. They cost seconds and catch most of what goes wrong.

• Units. Do the units of your expression come out as the units of the thing you wanted? m/s${}^2$ times s${}^2$ gives meters, so $\frac{1}{2}a{t}^2$ can be a displacement. If your answer for a distance has seconds in it, you have found an error without checking any arithmetic.
• Sign. Read the sign back as a direction, out loud if necessary. “Negative, so it’s below the launch point — yes, that’s the river.”
• Size. Is the number the right order of magnitude for a physical object? A car braking in 3 s, a ball rising 11 m, a marble landing 1 m from the table: plausible. A ball rising 1100 m is not.
• Bounds. An average must lie between the minimum and the maximum. A final speed after braking cannot exceed the initial. An angled projectile’s speed at the top cannot be zero — it still carries its full $v_x$. (Thrown straight up, like Problem 4, is the exception.)
• A second route. Where a problem can be done two ways, do the second one on the numbers you already have — it takes one line and it is the strongest check available. An answer you can defend two ways is an answer; an answer you can defend zero ways is a draft.