Kinematics.
The grammar of motion — how position, velocity, and acceleration describe every moving thing in the universe, from a dropped pencil to a probe crossing the solar system.
The physics of motion, from first principles
Imagine you're standing on a cliff. In one hand, you hold a bullet. In the other, a rifle pointed perfectly horizontal. At exactly the same instant, you drop the first bullet and fire the second. The rifle bullet screams out at 800 m/s while the dropped bullet just falls straight down. Both start at the same height. Which one hits the ground first?
Almost everyone gets this wrong on their first guess. The answer — verified in countless MythBusters-style experiments — is that they hit the ground at exactly the same time. The rifle bullet lands somewhere half a mile away, of course, but it hits the dirt at the same instant as the one you dropped at your feet.
This is the quiet power of kinematics: the branch of physics that describes how things move without yet worrying about why they move (that's next unit's job). Kinematics is the grammar of motion. Every problem in AP Physics 1 — every collision, every pendulum, every planet in orbit — eventually reduces to: where is the object, how fast is it going, and how is that speed changing? Master this unit, and you have the vocabulary to describe every physical situation you'll meet for the rest of the course.
Position, displacement, and the object model
Physics begins with a simplification so useful it feels like cheating: we replace real, complicated objects with idealized points. A sprinter is 6 feet tall, has muscles, arms, hair — but for the purposes of asking how far did she run, we pretend she's a single dot located at her center of mass. This is the object model, and it works because for translational motion, what matters is only where the object is as a whole, not its internal structure.
We track that dot with a position \(x\) — a single number that tells you where the object is relative to some chosen origin. The moment you draw an x-axis on your page, you have committed to a coordinate system. Rightward might be positive; leftward, negative. This sign convention is the single most important artistic choice you'll make on every problem.
Displacement \(\Delta x = x_f - x_0\) is how much your position changed. If you walk 5 meters east then 3 meters west, your displacement is \(+2\) m — a vector pointing east. But your distance traveled is \(5 + 3 = 8\) m. Distance is a scalar (just a number, always positive); displacement is a vector (number plus direction). Mixing these up is the single most common error on Unit 1 free-response questions.
Scalar quantities have magnitude only: mass, temperature, distance, speed, time. Vector quantities have magnitude and direction: position, displacement, velocity, acceleration, force. On the AP exam, if a problem asks for "the velocity" (vector), an answer without a direction is incomplete.
Velocity: the rate of change of position
The average velocity over an interval is displacement divided by the time it took:
This is a vector, in the same direction as the displacement. A car that drives from mile marker 100 to mile marker 50 in one hour has an average velocity of \(-50\) mph (negative because the displacement was negative). Its average speed, however, is the positive number \(50\) mph. Speed is a scalar; velocity is a vector.
Average velocity is useful, but it hides a lot. A car that accelerates from rest, cruises for a while, then brakes to a stop at the next intersection might have an average velocity of 30 mph — but that tells you nothing about whether the driver ever hit 60. What we really want is the instantaneous velocity: how fast the object is moving at one particular moment.
Mathematically, instantaneous velocity is what average velocity becomes in the limit of a vanishingly small time interval:
AP Physics 1 is algebra-based, so you're not required to compute derivatives — but you are required to read them off graphs. On a position-vs-time graph, instantaneous velocity is the slope of the tangent line at that instant. A steep upward slope means fast positive velocity; a flat spot means the object is momentarily at rest; a downward slope means negative velocity (moving in the \(-x\) direction).
Acceleration: the rate of change of velocity
Now do the exact same thing with velocity. Average acceleration:
And instantaneous acceleration is the slope of the tangent to a velocity-vs-time graph. An object is accelerating whenever its velocity is changing — either in magnitude or in direction. That second part trips people up. A car rounding a curve at a constant 30 mph is accelerating, because the direction of its velocity vector is changing. We'll exploit this idea heavily in Unit 5, but plant the flag now.
Negative acceleration does not mean "slowing down." It means the acceleration vector points in the \(-x\) direction. A ball falling downward with \(a = -10\) m/s² is speeding up. A car that is moving in the \(-x\) direction and has acceleration \(a = +2\) m/s² is slowing down. The rule: if \(v\) and \(a\) have the same sign, the object speeds up. If they have opposite signs, it slows down.
The three kinematic equations — and where they come from
If — and only if — the acceleration is constant, three equations link position, velocity, acceleration, and time. They are the workhorses of the entire unit. Let me show you where they come from so you never have to memorize them as isolated facts.
Equation 1 comes directly from the definition of average acceleration. If acceleration is constant, then average and instantaneous acceleration are the same thing, so \(a = (v - v_0)/t\). Solving for \(v\):
Equation 2 comes from the area under a velocity-vs-time graph. When acceleration is constant, the v-vs-t graph is a straight line, and the area underneath (which equals the displacement) is a trapezoid. A trapezoid's area is the average of the two parallel sides times the width:
If you substitute in \(v = v_0 + at\), you get:
Equation 3 is what you get when you eliminate \(t\) from the first two. It's the "timeless" equation — useful whenever a problem doesn't mention time:
Three equations. Five variables: \(x_0, x, v_0, v, a, t\) (six if you count initial position). Each equation is missing exactly one of them. That's your strategy for picking which to use: identify the variable the problem doesn't mention, and reach for the equation that doesn't include it.
The stopping car
A car traveling at 30 m/s slams on its brakes and decelerates uniformly at 6 m/s². How far does it travel before stopping?
Identify what's missing
Given: \(v_0 = 30\) m/s, \(v = 0\), \(a = -6\) m/s². Unknown: \(\Delta x\). Notice: time isn't mentioned and isn't asked for. Use the "timeless" equation Kin-3.
Plug and solve
$$v^2 = v_0^2 + 2a\Delta x \;\Rightarrow\; 0 = (30)^2 + 2(-6)\Delta x \;\Rightarrow\; \Delta x = \frac{900}{12} = 75 \text{ m}$$
The car travels 75 m before coming to rest.
Free fall: gravity is just a constant acceleration
Near the surface of Earth, any object in free fall — not touching anything, no air resistance — accelerates downward at \(g \approx 10\) m/s² (the CED uses 10 for clean arithmetic; 9.8 or 9.81 also receives full credit). This is true whether the object is moving up, down, sideways, or briefly at rest at the top of its arc. Gravity doesn't care what the object is doing; it just keeps pulling with the same constant acceleration.
If you drop a ball and a feather in a vacuum chamber, they land at the same time. Galileo knew this 400 years ago. Mass does not affect gravitational acceleration (we'll prove this in Unit 2 when we meet \(F = ma\) and \(F_g = mg\)).
The sign of \(g\) in your equations depends entirely on your coordinate choice. If you point the \(+y\) axis upward (the usual convention), then \(a = -g = -10\) m/s². If you point \(+y\) downward (sometimes useful for thrown-downward problems), then \(a = +g = +10\) m/s². Either works; just be consistent within a problem.
The dropped rock
A rock is dropped from rest off a cliff and hits the ground 3.0 s later. How tall is the cliff, and how fast is the rock moving at impact?
Choose a coordinate system
Let \(+y\) point upward, with the origin at the top of the cliff (so \(y_0 = 0\)). Then \(v_0 = 0\), \(a = -g = -10\) m/s², \(t = 3.0\) s.
Find position at impact
$$y = y_0 + v_0 t + \tfrac{1}{2}at^2 = 0 + 0 + \tfrac{1}{2}(-10)(3.0)^2 = -45 \text{ m}$$
The rock is 45 m below the origin, so the cliff is 45 m tall.
Find velocity at impact
$$v = v_0 + at = 0 + (-10)(3.0) = -30 \text{ m/s}$$
The magnitude is 30 m/s; the negative sign indicates downward motion, as expected.
Cliff height: 45 m · Impact speed: 30 m/s down
Graphical analysis — the secret language of the AP exam
AP Physics 1 loves graphs. Three-quarters of the FRQs will ask you to sketch, interpret, or extract numbers from a graph. The payoff of understanding the three kinematic graphs — \(x\)-vs-\(t\), \(v\)-vs-\(t\), and \(a\)-vs-\(t\) — is that you can often solve a problem faster graphically than algebraically.
Here is the structure of the relationship you must memorize:
| On a graph of… | The slope gives you… | The area under the curve gives you… |
|---|---|---|
| Position vs. time | Velocity | (nothing useful) |
| Velocity vs. time | Acceleration | Displacement |
| Acceleration vs. time | (nothing useful in AP 1) | Change in velocity |
So: if you're handed a \(v\)-vs-\(t\) graph and asked for the displacement between \(t = 2\) s and \(t = 5\) s, you don't plug into an equation — you count the area of the region under the curve between those times. Triangles, rectangles, trapezoids: break the shape into pieces with known areas.
Conversely, if a position graph curves upward, its slope is increasing — so velocity is increasing — so acceleration is positive. A parabolic-looking position graph opening upward is the signature of uniform positive acceleration. Opening downward signals uniform negative acceleration. A straight-line position graph means constant velocity (zero acceleration).
Reference frames and relative motion
Here is a deceptively deep idea: velocity is always measured relative to something. When you say a car is going 60 mph, you implicitly mean relative to the ground. To a passenger inside the car, the car's velocity is zero — but the ground is moving backward at 60 mph. Both observers are equally correct; neither is wrong.
This gets interesting when two things are both moving. If you're walking forward on a train at 2 m/s relative to the train, and the train is moving at 30 m/s relative to the ground, then your velocity relative to the ground is \(30 + 2 = 32\) m/s. If you'd walked toward the back, it'd be \(30 - 2 = 28\) m/s. The general rule, in one dimension:
In 2D this becomes a vector sum, which we'll formalize momentarily. The deep takeaway: there's no such thing as absolute velocity. Einstein built relativity out of this single observation.
Two dimensions: motion has two independent stories
Back to our opening puzzle: the dropped bullet and the fired bullet hit the ground at the same time. Why? Because horizontal and vertical motions are independent. The horizontal component of the fired bullet's velocity has no bearing whatsoever on how fast it falls. Gravity pulls only downward, so it only affects the vertical component.
When a 2D problem lands on your desk — a ball kicked off a cliff, a baseball thrown to the outfielder, a cannonball fired at an angle — your first move is always the same: break the initial velocity into its components.
Then you solve two completely separate 1D problems. The horizontal one has no acceleration (ignoring air resistance), so \(x = v_{0x}\,t\). The vertical one has \(a_y = -g\), so you apply the kinematic equations as usual. The two stories share only one variable — time — because time is the same in both directions.
Horizontal launch off a table
A ball rolls off a 1.25 m high table with a horizontal speed of 2.0 m/s. How far from the base of the table does it land?
Split into two problems
Vertical: \(y_0 = 1.25\) m, \(v_{0y} = 0\), \(a_y = -10\) m/s². Horizontal: \(v_{0x} = 2.0\) m/s, \(a_x = 0\).
Find time of flight (use the vertical story)
Let ground be \(y = 0\). Then \(0 = 1.25 + 0 + \tfrac{1}{2}(-10)t^2\), which gives \(t^2 = 0.25\), so \(t = 0.5\) s.
Find range (use the horizontal story with that time)
$$x = v_{0x} \, t = (2.0)(0.5) = 1.0 \text{ m}$$
The ball lands 1.0 m from the base of the table.
Where this unit takes you next
Everything in AP Physics 1 either is kinematics or is built on top of kinematics. In Unit 2, we'll ask what causes acceleration and meet Newton's laws. In Unit 3, we'll introduce energy, which is a shortcut that sometimes lets you bypass kinematics entirely. In Unit 4, momentum gives you another shortcut for collisions. In Unit 5, rotational kinematics is a carbon copy of this unit with different variable names (angles instead of positions). In Unit 7, oscillations are a specific kind of non-constant-acceleration motion that you'll handle with a small toolkit of sinusoidal shortcuts. The kinematic vocabulary — position, velocity, acceleration, vectors, graphs, coordinate choice — runs through every unit. Get it solid here; everything else gets easier.
Conceptual summary
Kinematics describes motion using position, velocity, and acceleration — all vectors, all defined relative to a chosen coordinate system. Average quantities use ratios of changes over intervals; instantaneous quantities are slopes of tangent lines on graphs. For constant acceleration, three equations relate these quantities; choosing which to use is a matter of spotting the variable the problem doesn't mention. Graphs encode the same information in a form the AP exam loves to test — slope gives the next quantity up (from \(x\) to \(v\), from \(v\) to \(a\)); area gives the next quantity down (from \(v\) to \(\Delta x\), from \(a\) to \(\Delta v\)). Horizontal and vertical motions are independent, which turns every projectile problem into two linked 1D problems sharing only a single time variable. Velocity is always relative to something, and gravity is just a constant downward acceleration of magnitude \(g \approx 10\) m/s² that treats all objects identically.
Every equation in Unit 1, at a glance
Stars (★) mark equations that appear on the AP Physics 1 equation sheet provided on exam day. Unstarred entries are definitions or derivations you're expected to know how to recover, but you won't see them printed on the exam.
Average velocity (definition)
- \(\bar v\)
- average velocity (m/s)
- \(\Delta x\)
- displacement (m)
- \(\Delta t\)
- time interval (s)
Use when you need a velocity averaged over an interval. Works for any motion (not just constant acceleration).
Average acceleration (definition)
- \(\bar a\)
- average acceleration (m/s²)
- \(\Delta v\)
- change in velocity (m/s)
- \(\Delta t\)
- time interval (s)
Use when you want a single acceleration value over an interval. Works for any motion.
★ Kinematic equation 1 — velocity / time
- \(v_x\)
- final velocity at time \(t\) (m/s)
- \(v_{x0}\)
- initial velocity (m/s)
- \(a_x\)
- constant acceleration (m/s²)
- \(t\)
- elapsed time (s)
Use when the problem doesn't involve position (\(x\)). Requires constant acceleration.
★ Kinematic equation 2 — position / time
- \(x\)
- position at time \(t\) (m)
- \(x_0\)
- initial position (m)
- \(v_{x0}\)
- initial velocity (m/s)
- \(a_x\)
- constant acceleration (m/s²)
- \(t\)
- elapsed time (s)
Use when the problem doesn't involve final velocity (\(v\)). Requires constant acceleration.
★ Kinematic equation 3 — "timeless"
- \(v_x\)
- final velocity (m/s)
- \(v_{x0}\)
- initial velocity (m/s)
- \(a_x\)
- constant acceleration (m/s²)
- \(x - x_0\)
- displacement (m)
Use when time (\(t\)) is neither given nor asked for. Requires constant acceleration.
Average velocity (constant-acceleration form)
- \(\bar v\)
- average velocity (m/s)
- \(v_0,\, v\)
- initial, final velocity (m/s)
Use when acceleration is constant. Derivable; not on the equation sheet but quick for simple problems.
Free-fall acceleration
- \(g\)
- gravitational acceleration near Earth's surface
Use when air resistance is negligible. CED specifies \(g = 10\) for clean arithmetic, though 9.8 or 9.81 m/s² is also accepted. Sign depends on your choice of axis: if up is positive, \(a_y = -g\).
Vector components
- \(v\)
- magnitude of the vector (m/s)
- \(\theta\)
- angle measured from the \(+x\)-axis (degrees or radians)
Use when you need to decompose a 2D velocity (or any vector) into components. First step of every projectile problem.
Vector magnitude from components
- \(v\)
- total speed (m/s)
- \(\theta\)
- angle from \(+x\)-axis
Use when reassembling a vector from its components — e.g., finding the speed and angle of a projectile at some moment.
Relative velocity (1D)
- Subscripts
- read as "velocity of A relative to C"
Use when multiple reference frames are in play (passenger on train, boat in river, plane in wind). In 2D this becomes a vector sum.
Every answer on the AP exam must carry units. Position: m. Velocity: m/s. Acceleration: m/s². Time: s. If you get an answer in cm, convert — physics equations all assume SI units. A surprising number of free-response points are lost to unitless final answers.
One page. Night before the test.
The 5 concepts that matter most
- Vectors carry direction; scalars don't. Displacement, velocity, and acceleration are vectors — a sign (or direction) is part of the answer, not decoration. Distance and speed are scalars.
- "Constant acceleration" is the magic phrase. The three kinematic equations work only when \(a\) is constant. If acceleration changes, you need graphs (slopes and areas) instead.
- Negative acceleration ≠ slowing down. Signs of \(v\) and \(a\) together determine speeding up (same sign) vs slowing down (opposite signs).
- Slopes and areas. On an \(x\)-vs-\(t\) graph, slope = velocity. On a \(v\)-vs-\(t\) graph, slope = acceleration and area = displacement. On an \(a\)-vs-\(t\) graph, area = change in velocity.
- 2D motion = two 1D problems linked by time. In projectile motion, \(a_x = 0\) and \(a_y = -g\). The horizontal and vertical stories share only \(t\).
Key equations
- \(\bar v = \Delta x / \Delta t\)
- \(\bar a = \Delta v / \Delta t\)
- \(v = v_0 + at\)
- \(x = x_0 + v_0 t + \tfrac{1}{2}a t^2\)
- \(v^2 = v_0^2 + 2a(x - x_0)\)
- \(v_x = v\cos\theta, \; v_y = v\sin\theta\)
- \(g \approx 10\) m/s² (down)
5 traps students fall for
- Using a kinematic equation when acceleration isn't constant.
- Forgetting \(g\) is negative if \(+y\) points up.
- Assuming the object is at rest at the highest point means \(a = 0\) there (it doesn't — \(a = -g\) the whole time).
- Adding speeds of two things moving in 2D as if they were 1D (they're vectors).
- Reporting distance instead of displacement, or vice versa.
Problem-solving checklist
- Draw a picture and a coordinate axis. Pick which direction is positive. Stick with it.
- List what you know and what you want. Write them with symbols, not just numbers (e.g., "\(v_0 = +15\) m/s").
- Check: is acceleration constant? If yes → kinematic equations. If no → graphs.
- Identify the missing variable. Choose the kinematic equation that doesn't contain it.
- Solve symbolically first, then substitute numbers. (AP FRQs often demand symbolic answers.)
- Check units and signs. Does the sign of your answer point the way physical intuition says it should?
Ten problems, scaffolded from basic to moderate
A cyclist rides 300 m east, then 100 m west, all in 40 s. Calculate (a) her average speed and (b) her average velocity (magnitude and direction).
A car accelerates uniformly from rest to 20 m/s in 5.0 s. What is its acceleration?
A ball is dropped from rest off the top of a building and falls for 2.0 s before hitting the ground. How tall is the building? (Use \(g = 10\) m/s².)
A car traveling at 25 m/s brakes uniformly and comes to a stop over a distance of 50 m. (a) What is the magnitude of its deceleration? (b) How long did it take to stop?
A rock is thrown straight up with an initial speed of 20 m/s from ground level. (a) How high does it rise? (b) How long is it in the air before returning to the thrower's hand?
A runner accelerates uniformly from 3.0 m/s to 9.0 m/s over a distance of 18 m. (a) What is her acceleration? (b) How long did the sprint take?
A ball is thrown horizontally off a cliff with an initial speed of 15 m/s. It hits the ground 3.0 s later. (a) How tall is the cliff? (b) How far from the base of the cliff does it land? (Ignore air resistance.)
The velocity of an object moving along the \(x\)-axis is given by the graph below:
(a) What is the object's acceleration from \(t = 0\) to \(t = 2\) s? (b) What is the total displacement from \(t = 0\) to \(t = 10\) s?
A ball is thrown straight up with initial speed \(v_0\) from the edge of a 20 m tall building. It hits the ground below 4.0 s after being released. Find \(v_0\).
A projectile is launched from ground level at 30 m/s at an angle of 37° above the horizontal. Using \(\sin 37° \approx 0.6\), \(\cos 37° \approx 0.8\), and \(g = 10\) m/s², find (a) the maximum height, (b) the total time of flight, and (c) the horizontal range.
Ten problems in the exact format of the real AP exam
Part I — Multiple Choice
A ball is thrown straight upward. At the instant it reaches its highest point, which of the following is true?
- The velocity is zero and the acceleration is zero.
- The velocity is zero and the acceleration points downward.
- The velocity points downward and the acceleration is zero.
- The velocity and acceleration both point downward.
Two cars, A and B, start from rest at the same point at \(t = 0\) and accelerate along a straight road. Car A has acceleration \(2a\) and car B has acceleration \(a\). At time \(t\), what is the ratio of car A's displacement to car B's displacement?
- 1 : 1
- \(\sqrt{2}\) : 1
- 2 : 1
- 4 : 1
The position of an object moving along the \(x\)-axis as a function of time is shown. Which of the following best describes the object's velocity?
- Velocity is constant and positive.
- Velocity is positive and increasing in magnitude.
- Velocity is positive and decreasing in magnitude.
- Velocity is negative and increasing in magnitude.
A student drops a ball from a window of height \(h\). A second ball is dropped from a height \(4h\) above the ground. Ignoring air resistance, the ratio of the time it takes the second ball to reach the ground to the time it takes the first is:
- 1 : 1
- \(\sqrt{2}\) : 1
- 2 : 1
- 4 : 1
An object moves along the \(x\)-axis with the velocity-vs-time graph shown. Over the interval from \(t = 0\) to \(t = 8\) s, which statement is correct?
- The object is speeding up the entire time.
- The object is slowing down from \(t = 0\) to 4 s, then speeding up from 4 s to 8 s.
- The object has zero displacement because it reverses direction.
- The object has constant acceleration that changes sign at \(t = 4\) s.
Part II — Short Free Response
(Translation between representations, ~7 minutes) An object moves along a straight horizontal track. Its velocity as a function of time is given by the graph shown, where positive values indicate motion in the \(+x\) direction.
- Describe, in words, the motion of the object during each of the three intervals: 0 to 2 s, 2 to 4 s, and 4 to 6 s. Discuss direction, speeding up vs. slowing down, and acceleration for each interval.
- Calculate the net displacement of the object from \(t = 0\) to \(t = 6\) s. Show your reasoning using the area under the curve.
(Qualitative/quantitative translation, ~7 minutes) A student uses a motion sensor to record the position of a glider on a frictionless air track as it is given a brief push and then slides freely. The push ends at \(t = 0\), and the glider moves in the \(+x\) direction.
- Sketch a qualitatively correct graph of position vs time for the glider from \(t = 0\) to the moment it collides elastically with a wall and reverses direction. Label the axes.
- Explain, in terms of velocity and acceleration, why your graph has the shape it does during the free-sliding phase.
- A second student claims that because the glider is moving at constant velocity, the acceleration is also constant (but nonzero). Evaluate this claim in one to two sentences.
(Mathematical routines, ~7 minutes) A student on a bridge drops a stone from rest. The stone falls for a time \(T\) before hitting the water below. Neglect air resistance.
- Derive a symbolic expression for the height \(H\) of the bridge above the water in terms of \(T\) and physical constants. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
- Derive a symbolic expression for the speed \(v\) of the stone the instant before it hits the water in terms of \(T\) and physical constants.
- The student now drops a second stone of twice the mass from the same bridge. Without redoing the calculation, indicate whether \(T\) is greater than, less than, or equal to the original fall time, and justify your answer.
Part III — Extended Free Response
(Experimental design and analysis, ~18 minutes) A student is asked to determine the acceleration due to gravity \(g\) in a laboratory. The student has a meter stick, an electronic timer that records the time a photogate beam is broken, and a steel ball that can be released from various heights \(h\) above a photogate.
- Describe an experimental procedure the student could follow to determine \(g\). Include the quantities the student should measure and vary, and how multiple trials would be used.
- The student drops the ball from various heights \(h\) and records the speed \(v\) at which the ball crosses the photogate (calculated from a known ball diameter and the photogate-break time). Explain how a graph of \(v^2\) vs \(h\) could be used to determine \(g\), including what the expected slope should be.
- The student's data yields a best-fit line of slope \(19.6\) m/s² per meter. Calculate the experimental value of \(g\).
- The student notices that all of her \(v^2\) data points lie slightly below the best-fit line at large \(h\). Propose a physical explanation for this systematic trend.
(Mathematical routines + translation, ~18 minutes) A cannon fires a projectile from ground level with initial speed \(v_0\) at an angle \(\theta\) above the horizontal. The projectile lands back on level ground. Assume no air resistance.
- Derive a symbolic expression for the total time of flight \(T\) in terms of \(v_0\), \(\theta\), and physical constants. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
- Derive a symbolic expression for the horizontal range \(R\) in terms of \(v_0\), \(\theta\), and physical constants.
- For a fixed \(v_0\), derive the angle \(\theta\) that maximizes \(R\), and find the maximum range.
- Using the expression from part (b), indicate whether launch angles of 30° and 60° produce the same range, a larger range for 30°, or a larger range for 60°. Justify your answer using your expression for \(R\).
- Sketch a qualitatively correct graph of the projectile's vertical velocity \(v_y\) as a function of time for the entire flight. Label the axes and include the values \(v_y(0)\), \(v_y(T/2)\), and \(v_y(T)\) in terms of \(v_0\) and \(\theta\).
Five problems at F=ma / olympiad difficulty
The falling meter stick. A 1.0 m long stick is held vertically just above the floor and released from rest. Simultaneously, a small ball is dropped from rest from a height of 1.0 m above the floor. Each falls without air resistance. Which reaches the floor first, or do they tie? Justify your answer quantitatively — compute the time for each using energy conservation or kinematic reasoning about the bottom of each object, and explain the physical reason for any difference.
Two-car chase (derivation required). Car A, traveling with constant velocity \(v\), passes a stationary police car B at \(t = 0\). The police car accelerates uniformly from rest at acceleration \(a\) starting at \(t = 0\) and pursues. Derive symbolic expressions for:
- the time \(t^\ast\) at which B catches up to A,
- the distance \(d^\ast\) B has traveled at that moment,
- B's speed at the catch-up instant, and show this speed is exactly \(2v\) — then explain in one sentence why this must be true using the idea of average velocity.
Graphical analysis (no equations allowed). The acceleration of a particle moving along the \(x\)-axis is shown below as a function of time. At \(t = 0\), the particle is at \(x_0 = 0\) with \(v_0 = +4\) m/s.
Using only the area-under-the-curve rule (no kinematic equations), determine the particle's velocity at \(t = 10\) s. Then, by reasoning about the resulting velocity graph, determine whether the particle is at \(x > 0\) or \(x < 0\) at \(t = 10\) s, and justify your answer.
Optimal throwing angle off a cliff. A ball is thrown with speed \(v_0\) from the edge of a cliff of height \(h\) above level ground. The launch angle \(\theta\) above the horizontal can be chosen to maximize the horizontal range (where the ball lands on the ground below). Show that the optimal angle satisfies
$$\tan^2\theta_{\text{opt}} = \dfrac{v_0^2}{v_0^2 + 2gh}$$
and verify that this recovers the familiar result \(\theta_{\text{opt}} = 45°\) when \(h = 0\). Explain physically why \(\theta_{\text{opt}} < 45°\) whenever \(h > 0\).
The monkey and the hunter. A hunter aims a rifle directly at a monkey hanging from a tree branch a horizontal distance \(d\) away at a height \(h\) above the hunter's rifle. At the instant the rifle fires, the monkey releases its grip and begins to fall. Assume the bullet leaves the rifle with speed \(v_0\) and neglect air resistance.
- Prove that the bullet always hits the monkey, regardless of \(v_0\), provided \(v_0\) is large enough to reach the tree before the monkey hits the ground.
- Find the minimum speed \(v_{0,\text{min}}\) such that the bullet reaches the tree before the monkey hits the ground. Express your answer in terms of \(d\), \(h\), and physical constants.
Every problem, every step, every why
Worksheet A — Foundation
(a) Average speed. Total distance traveled is \(300 + 100 = 400\) m. Time is 40 s. Average speed = distance / time = \(400 / 40 = 10\) m/s.
(b) Average velocity. Net displacement: 300 m east − 100 m east = \(+200\) m. Average velocity = displacement / time = \(+200/40 = +5\) m/s.
Given \(v_0 = 0\), \(v = 20\) m/s, \(t = 5.0\) s. Use \(a = \Delta v / \Delta t\):
$$a = \frac{20 - 0}{5.0} = 4.0 \text{ m/s}^2$$
Take downward as positive. Then \(v_0 = 0\), \(a = +10\) m/s², \(t = 2.0\) s. Using \(x = x_0 + v_0 t + \tfrac{1}{2}a t^2\) with \(x_0 = 0\):
$$x = 0 + 0 + \tfrac{1}{2}(10)(2.0)^2 = 20 \text{ m}$$
Given \(v_0 = 25\) m/s, \(v = 0\), \(\Delta x = 50\) m. Time isn't given, so start with the timeless equation \(v^2 = v_0^2 + 2a\Delta x\):
$$0 = (25)^2 + 2a(50) \;\Rightarrow\; a = -\frac{625}{100} = -6.25 \text{ m/s}^2$$
Magnitude of deceleration: 6.25 m/s². Then \(t = (v - v_0)/a = (0 - 25)/(-6.25) = 4.0\) s.
Take up as positive. Then \(v_0 = +20\) m/s, \(a = -10\) m/s². At the top, \(v = 0\).
(a) Maximum height — use timeless equation:
$$0 = (20)^2 + 2(-10)h \;\Rightarrow\; h = \frac{400}{20} = 20 \text{ m}$$
(b) Total time — by symmetry, time up = time down. Time to top: \(0 = 20 + (-10)t \Rightarrow t = 2.0\) s. Total: \(2 \times 2.0 = 4.0\) s.
Given \(v_0 = 3.0\) m/s, \(v = 9.0\) m/s, \(\Delta x = 18\) m. Time not given.
(a) \(v^2 = v_0^2 + 2a\Delta x\): \((9)^2 = (3)^2 + 2a(18) \Rightarrow 81 = 9 + 36a \Rightarrow a = 2.0\) m/s².
(b) \(t = (v - v_0)/a = 6/2 = 3.0\) s.
Split into components. Horizontal: \(v_{0x} = 15\) m/s, \(a_x = 0\). Vertical: \(v_{0y} = 0\), \(a_y = -10\) m/s² (up positive). Time of flight: \(t = 3.0\) s.
(a) Cliff height. Take the top of the cliff as \(y = 0\). At \(t = 3.0\) s: \(y = 0 + 0 + \tfrac{1}{2}(-10)(3)^2 = -45\) m, so the cliff is 45 m tall.
(b) Range. \(x = v_{0x}\,t = (15)(3.0) = 45\) m.
(a) Acceleration 0–2 s. Velocity rises from 2 to 8 m/s over 2 s, so slope = \((8-2)/2 = +3.0\) m/s².
(b) Total displacement 0–10 s. Break the area under the v–t graph into pieces:
- 0 to 2 s: trapezoid, \(\tfrac{1}{2}(2 + 8)(2) = 10\) m
- 2 to 6 s: rectangle, \((8)(4) = 32\) m
- 6 to 10 s: triangle, \(\tfrac{1}{2}(8)(4) = 16\) m
Total: \(10 + 32 + 16 = 58\) m.
Let up be positive. Take the top of the building as \(y = 0\); the ground is at \(y = -20\) m. Given \(a = -10\), \(t = 4.0\) s, \(y = -20\), find \(v_0\). Use \(y = v_0 t + \tfrac{1}{2}a t^2\):
$$-20 = v_0(4.0) + \tfrac{1}{2}(-10)(4.0)^2 = 4v_0 - 80 \;\Rightarrow\; v_0 = \frac{60}{4} = 15 \text{ m/s}$$
Components: \(v_{0x} = 30 \cos 37° = 24\) m/s; \(v_{0y} = 30 \sin 37° = 18\) m/s.
(a) Max height. At peak, \(v_y = 0\). Using \(v_y^2 = v_{0y}^2 + 2(-g)h\): \(0 = (18)^2 - 20h \Rightarrow h = 324/20 = 16.2\) m.
(b) Time of flight. Using \(y = v_{0y}t + \tfrac{1}{2}(-g)t^2 = 0\) (lands at same height): \(t(18 - 5t) = 0\), so \(t = 3.6\) s.
(c) Range. \(R = v_{0x}\,t = (24)(3.6) = 86.4\) m.
Worksheet B — AP Exam Style
At the top of a ball's flight, the velocity is momentarily zero — but gravity is still acting, and acceleration is still \(-g\) downward. Acceleration does not care what the velocity is; gravity pulls on a ball the same way whether it's rising, falling, or instantaneously at rest.
From rest with acceleration \(\alpha\) for time \(t\), displacement is \(\Delta x = \tfrac{1}{2}\alpha t^2\). So car A: \(\Delta x_A = \tfrac{1}{2}(2a)t^2 = a t^2\). Car B: \(\Delta x_B = \tfrac{1}{2}a t^2\). Ratio: \(\Delta x_A / \Delta x_B = 2\).
The curve starts with shallow positive slope (small positive velocity) and becomes steeper as time goes on (larger positive velocity). So velocity is positive and increasing.
For an object dropped from rest, \(h = \tfrac{1}{2}g t^2\), so \(t = \sqrt{2h/g}\). If heights are in ratio 4 : 1, times are in ratio \(\sqrt{4} : \sqrt{1} = 2 : 1\).
The graph is a straight line sloping downward through \(v = 0\) at \(t = 4\) s. That means acceleration (the slope) is constant and negative the entire time — it does not change sign. From 0–4 s, \(v > 0\) and \(a < 0\): opposite signs, so the object is slowing down. From 4–8 s, \(v < 0\) and \(a < 0\): same sign, so the object is speeding up (in the negative direction).
(a) Interval descriptions.
- 0–2 s: \(v = +4\) m/s (constant, positive). Object moves in \(+x\) direction at constant speed. Acceleration = 0.
- 2–4 s: \(v\) decreases from \(+4\) to 0. Object is still moving in \(+x\) direction but slowing down. Acceleration is negative (slope of line is \(-2\) m/s²).
- 4–6 s: \(v\) decreases from 0 to \(-4\) m/s. Object reverses direction and now moves in \(-x\) direction, speeding up. Acceleration remains \(-2\) m/s² (same slope continues).
(b) Net displacement. Area above the \(t\)-axis (positive): rectangle from 0–2 s plus triangle from 2–4 s: \((4)(2) + \tfrac{1}{2}(4)(2) = 8 + 4 = 12\) m. Area below the axis (negative): triangle from 4–6 s: \(-\tfrac{1}{2}(4)(2) = -4\) m. Net displacement: \(12 - 4 = +8\) m.
(a) Position-time graph. After the push ends at \(t = 0\), the glider moves with constant velocity until it hits the wall. A straight line with positive slope. At the collision, it reverses instantaneously (assumed elastic) and a straight line with equal-magnitude negative slope begins.
(b) Explanation. During the free-sliding phase, no net force acts on the glider (frictionless track, no external forces), so acceleration is zero. Constant acceleration of zero means constant velocity, which on a position graph is a straight line.
(c) Evaluating the claim. The claim is incorrect. If velocity is truly constant, then \(\Delta v / \Delta t = 0\), so acceleration is zero — not a nonzero constant. A constant nonzero acceleration would cause the velocity to change linearly over time, contradicting "constant velocity."
(a) Height. Starting from \(y = y_0 + v_0 t + \tfrac{1}{2}a t^2\) with down-positive convention, \(y_0 = 0\), \(v_0 = 0\), \(a = g\), \(t = T\):
$$H = \tfrac{1}{2} g T^2$$
(b) Impact speed. From \(v = v_0 + a t\) with \(v_0 = 0\):
$$v = gT$$
(c) Mass-doubled stone. The fall time is equal to the original. Free-fall acceleration near Earth's surface is independent of mass; both stones experience \(a = g\) downward. Galileo's observation.
(a) Procedure. Measure the diameter \(d\) of the ball. Release the ball from rest from a known height \(h\) above the photogate. Measure the time \(\Delta t\) for the ball to break the beam. Compute the speed \(v = d/\Delta t\) as it crosses the gate. Repeat for multiple heights \(h\) to generate a data set. Take multiple trials at each height to reduce random error.
(b) Linearization. For a ball dropped from rest, the timeless kinematic equation gives \(v^2 = 2gh\). Plotting \(v^2\) (vertical) vs \(h\) (horizontal) should yield a straight line through the origin with slope \(2g\).
(c) Calculation. Slope = \(2g\) = 19.6 m/s² per m, so \(g = 9.8\) m/s².
(d) Systematic deviation at large \(h\). At large heights, the ball is traveling faster as it crosses the photogate, so air resistance (which grows with speed) removes more kinetic energy. This makes the measured \(v^2\) slightly smaller than the frictionless prediction, pulling data points below the best-fit line.
Components: \(v_{0x} = v_0 \cos\theta\), \(v_{0y} = v_0 \sin\theta\).
(a) Time of flight. Lands at \(y = 0\). From \(y = v_{0y}t - \tfrac{1}{2}gt^2 = 0\): \(t(v_{0y} - \tfrac{1}{2}g t) = 0\), so \(t = 0\) (launch) or \(t = 2v_{0y}/g\). Thus
$$T = \dfrac{2 v_0 \sin\theta}{g}$$
(b) Range. \(R = v_{0x}\,T = v_0\cos\theta \cdot \dfrac{2 v_0 \sin\theta}{g} = \dfrac{v_0^2 \sin(2\theta)}{g}\) using \(2\sin\theta\cos\theta = \sin(2\theta)\).
(c) Maximum range. \(\sin(2\theta)\) is maximized at \(2\theta = 90°\), so \(\theta = 45°\). Max range \(R_{\max} = v_0^2/g\).
(d) 30° vs 60°. \(\sin(60°) = \sin(120°) \approx 0.866\) — they're equal. So 30° and 60° produce the same range (complementary angles give equal ranges on level ground).
(e) \(v_y\) vs \(t\). Straight line starting at \(+v_0\sin\theta\) at \(t = 0\), crossing zero at \(t = T/2\), reaching \(-v_0\sin\theta\) at \(t = T\). Slope throughout is \(-g\).
Worksheet C — Challenge
Interpret "reaches the floor" as "any part of the object touches the floor." The stick is held vertically, so its bottom is already at the floor at \(t = 0\) — it reaches the floor immediately. But if we interpret the problem as "when does each object's center of mass arrive at the floor":
The ball is a point object; its center starts at height 1.0 m and falls freely: \(t_1 = \sqrt{2(1.0)/10} = \sqrt{0.2} \approx 0.45\) s.
The stick's center of mass starts at height 0.5 m (middle of the 1.0 m stick held with bottom at the floor). If it falls freely (which a rigid stick rotating about its bottom edge does NOT — but a stick simply released from vertical rest is translational free fall, not rotational), \(t_2 = \sqrt{2(0.5)/10} = \sqrt{0.1} \approx 0.32\) s.
However, the usual intent of this problem is that the stick is balanced vertically and then allowed to fall over, rotating about its bottom contact point. A uniform stick rotating about its end has angular acceleration \(\alpha = \tfrac{3g}{2L}\cos\theta\) where \(\theta\) is the angle from vertical. The tip follows an arc, and its vertical free-fall time (if dropped point-like from the tip's height of 1.0 m) would be \(\sqrt{0.2} \approx 0.45\) s. But at small angles the tip falls faster than free fall because the angular acceleration builds quickly — so the tip of the stick actually hits the floor before a point-mass dropped from the same height.
Car A: \(x_A(t) = v t\). Car B (from rest, acceleration \(a\)): \(x_B(t) = \tfrac{1}{2} a t^2\).
(a) Catch-up time. Set equal: \(v t^\ast = \tfrac{1}{2} a (t^\ast)^2\). Dividing by \(t^\ast\) (excluding the \(t = 0\) solution): \(v = \tfrac{1}{2} a t^\ast \Rightarrow\)
$$t^\ast = \dfrac{2v}{a}$$
(b) Distance. \(d^\ast = v \cdot t^\ast = \dfrac{2v^2}{a}\).
(c) Speed of B at catch-up. \(v_B(t^\ast) = a t^\ast = a \cdot \dfrac{2v}{a} = 2v\). Why this must be true: B starts from rest and accelerates uniformly, so its average velocity during the pursuit is \(\tfrac{1}{2}v_B(t^\ast)\). For B to cover the same distance as A in the same time, the average velocities must match: \(\tfrac{1}{2}v_B(t^\ast) = v\), hence \(v_B(t^\ast) = 2v\).
Velocity at \(t = 10\) s. \(\Delta v\) = area under the \(a\)–\(t\) graph. Reading the piecewise acceleration:
- 0–2 s: \(a = +2\), rectangle area \(= (2)(2) = +4\) m/s
- 2–6 s: \(a = -2\), area \(= (-2)(4) = -8\) m/s
- 6–10 s: \(a = 0\), area \(= 0\)
Net \(\Delta v = +4 - 8 + 0 = -4\) m/s. Starting from \(v_0 = +4\) m/s: \(v(10) = +4 + (-4) = 0\) m/s.
Position at \(t = 10\) s. Sketch the \(v\)–\(t\) graph that results from this acceleration:
- 0–2 s: \(v\) rises linearly from \(+4\) to \(+8\) m/s
- 2–6 s: \(v\) falls linearly from \(+8\) to 0 m/s (crosses zero somewhere in this interval)
- 6–10 s: \(v = 0\) (rests)
When does \(v\) cross zero? From \(v = 8 + (-2)(t-2) = 0\): \(t = 6\) s exactly. So the particle moves in \(+x\) the entire interval 0–6 s and then rests. Displacement = area under positive \(v\)–\(t\) graph:
- 0–2 s: trapezoid, \(\tfrac{1}{2}(4+8)(2) = 12\) m
- 2–6 s: triangle, \(\tfrac{1}{2}(8)(4) = 16\) m
- 6–10 s: 0 m
Total: \(+28\) m. Since \(x_0 = 0\), the particle is at \(x = +28\) m, so \(x > 0\).
Launch from height \(h\) with speed \(v_0\) at angle \(\theta\). Vertical: \(y = h + v_0\sin\theta\,t - \tfrac{1}{2}g t^2\). Setting \(y = 0\) at landing gives a quadratic in \(t\). Take the positive root:
$$t = \frac{v_0\sin\theta + \sqrt{v_0^2\sin^2\theta + 2gh}}{g}$$
Range: \(R(\theta) = v_0\cos\theta \cdot t\). To maximize, differentiate \(R\) with respect to \(\theta\) and set equal to zero. After a fair bit of algebra (the calculation involves setting \(dR/d\theta = 0\) and solving the resulting equation), one can show that at the optimum:
$$\tan^2\theta_{\text{opt}} = \dfrac{v_0^2}{v_0^2 + 2gh}$$
When \(h = 0\): \(\tan^2\theta_{\text{opt}} = 1 \Rightarrow \tan\theta = 1 \Rightarrow \theta = 45°\), confirming the classical result.
Physical explanation. When launching from a height, you already have extra "falling time" to work with — the projectile spends more time in the air regardless of launch angle than it would from ground level. To exploit this, you want to bias the initial velocity toward the horizontal component (to cover more distance per unit time), which means \(\theta < 45°\).
Let the hunter be at origin, the monkey at horizontal position \(d\) and height \(h\). The rifle is aimed directly at the monkey, so the launch angle satisfies \(\tan\theta = h/d\), and the components are \(v_{0x} = v_0\cos\theta\), \(v_{0y} = v_0\sin\theta\).
(a) Do they collide? At time \(t\), the bullet's position is:
$$x_{\text{bullet}}(t) = v_0\cos\theta \cdot t \qquad y_{\text{bullet}}(t) = v_0\sin\theta \cdot t - \tfrac{1}{2}g t^2$$
Meanwhile, the monkey drops from \((d, h)\) with initial velocity zero: \(y_{\text{monkey}}(t) = h - \tfrac{1}{2}g t^2\). The bullet reaches horizontal position \(d\) at time \(t^\ast = d/(v_0\cos\theta)\). At that moment:
$$y_{\text{bullet}}(t^\ast) = v_0\sin\theta \cdot \dfrac{d}{v_0\cos\theta} - \tfrac{1}{2}g(t^\ast)^2 = d\tan\theta - \tfrac{1}{2}g(t^\ast)^2$$
Using \(\tan\theta = h/d\): \(y_{\text{bullet}}(t^\ast) = h - \tfrac{1}{2}g(t^\ast)^2\). And at the same moment: \(y_{\text{monkey}}(t^\ast) = h - \tfrac{1}{2}g(t^\ast)^2\). They're equal, always. The bullet hits the monkey regardless of \(v_0\).
Why? Both are in free fall starting at the same instant. In the absence of gravity, the bullet would travel in a straight line from the rifle to the monkey's original position. Gravity pulls both the bullet and the monkey down by the same amount \(\tfrac{1}{2}g t^2\) at every instant. So the bullet "drops" relative to its straight-line path by exactly the same amount the monkey falls from its original position. Collision is guaranteed.
(b) Minimum speed. The monkey reaches the ground when \(\tfrac{1}{2}g t^2 = h\), i.e., \(t_{\text{ground}} = \sqrt{2h/g}\). For the bullet to reach the tree before this, we need \(t^\ast < t_{\text{ground}}\):
$$\dfrac{d}{v_0 \cos\theta} < \sqrt{\dfrac{2h}{g}}$$
Using \(\cos\theta = d/\sqrt{d^2+h^2}\):
$$v_0 > \dfrac{\sqrt{d^2+h^2}}{\sqrt{2h/g}} = \sqrt{\dfrac{g(d^2+h^2)}{2h}}$$
Hence \(v_{0,\text{min}} = \sqrt{g(d^2+h^2)/(2h)}\).
Eight errors that cost students points on this unit
Confusing distance with displacement
A cyclist rides 3 km north, then 4 km east. What is her displacement?
It feels natural because "distance traveled" and "displacement" sound like synonyms in ordinary English. Physics uses them with surgical precision.
An object moves 10 m east, then 6 m west, then 4 m east. What is the total distance traveled and the displacement?
Thinking \(a = 0\) at the top of a toss
A ball is thrown straight up. At the highest point, what is the acceleration?
Students conflate velocity with acceleration. The velocity is instantaneously zero at the peak, but that's precisely because gravity is constantly decelerating the ball on the way up and then accelerating it on the way down. If \(a\) were truly zero at the peak, the ball would float there forever.
A rock is thrown upward from ground level at 15 m/s. What is the rock's velocity 0.5 s after the peak? What is its acceleration at the peak, on the way up, and on the way down?
Using kinematic equations when \(a\) is not constant
A velocity-vs-time graph shows a curve that is clearly not a straight line. Find the displacement over a given interval.
The kinematic equations are so heavily drilled that students reach for them reflexively. But when the AP exam gives you a curved v-t graph, it's explicitly testing whether you know that your standard tools don't apply.
The velocity of a particle is given by \(v(t) = 3t\) m/s (not constant acceleration in magnitude — but actually constant with respect to time, so this is a trick; acceleration = 3 m/s². Now try \(v(t) = t^2\) m/s. Find the displacement over 0 to 2 s by finding the area under the v-t curve.)
Sign errors because the axis convention was never declared
A ball is thrown upward at 20 m/s. Find its velocity 3 s later.
Every sign error I've ever graded has the same root cause: the student forgot to pick an axis direction before plugging in numbers. Make this the first line of every problem you write.
A rock is dropped from a 45 m cliff. Taking downward as positive, what are the values (with sign) of \(v_0\), \(a\), \(y_0\), and \(y\) at impact? Now redo with upward as positive. The answer for how long it takes to hit the ground should be the same either way.
Adding 2D velocities as if they were scalars
A boat has engine speed 4 m/s in still water and heads straight across a river that flows east at 3 m/s. What is the boat's speed relative to the ground?
"Relative velocity" problems whisper "just add the numbers" because the simpler version is 1D addition. The moment velocities are in different directions, addition becomes vector addition — magnitude and direction, not just magnitude.
A plane flies northwest at 200 m/s relative to still air. A wind blows east at 50 m/s. What is the plane's velocity relative to the ground?
Assuming "negative acceleration" = "slowing down"
A car has acceleration \(-2\) m/s². Is it speeding up or slowing down?
English reinforces the misconception: we use "accelerate" to mean "speed up" and "decelerate" to mean "slow down." Physics doesn't. "Acceleration" is just a signed vector. Pair it with velocity's sign to determine what's happening.
A ball is dropped from rest. Take up as positive. For the entire fall, what are the signs of \(v\) and \(a\)? Is the ball speeding up or slowing down? Now throw the same ball upward at 10 m/s. On the way up, what are the signs? After the peak?
Not splitting projectile motion into two 1D problems
A ball is kicked off a cliff at 15 m/s at 37° above horizontal. How long until it hits the ground 20 m below?
Students see the total velocity number and reach for it. But only the vertical component matters for how fast the ball falls; only the horizontal component matters for how fast it goes sideways. These are two independent motions sharing a single time variable.
A projectile is launched at 25 m/s at 53° above horizontal from ground level. What are \(v_{0x}\) and \(v_{0y}\)? How long until the projectile returns to launch height?
Reading \(v_0 = 0\) as "starts at the origin"
A ball is thrown from the top of a 40 m building. It leaves the hand with speed 5 m/s upward. Find the height 2 s later.
The subscript-zero notation conflates two different quantities in students' minds. \(v_0\) means "initial velocity." \(x_0\) means "initial position." They are independent. An object can start at a nonzero position with zero velocity, or vice versa, or both nonzero.
A stone is dropped from the top of a 125 m cliff. Set up the kinematic equation for its position as a function of time, being careful to specify \(y_0\), \(v_0\), and \(a\) separately. When does it hit the ground?
What I learned going through this unit — and from helping others through it
What every student struggles with (in order)
The #1 sticking point in Unit 1 is not the equations — it's sign conventions. Students who have memorized all three kinematic equations can still get a projectile problem wrong because they didn't declare which direction is positive. The fix is boring but bulletproof: on every single problem, the first thing you write on the page is a coordinate axis and the direction you've labeled as positive. Do this even for problems you could solve in your head.
The #2 struggle is knowing when the kinematic equations don't apply. The three equations are so beautiful and so dependable that students reach for them whenever they see a motion problem. But if the acceleration isn't constant, all three equations give nonsense. Look for phrases like "uniform acceleration," "constant deceleration," or "free fall with no air resistance" — those are the problems where kinematic equations work. If the problem says "variable acceleration" or gives you a non-linear v-t graph, reach for areas and slopes instead.
What separates a 4 from a 5
Fluency with graphical analysis. Students who score a 5 can look at a v-t graph and immediately tell you: acceleration is the slope, displacement is the area under the curve, the object reverses direction wherever the curve crosses zero, and speeding-up-vs-slowing-down depends on whether \(v\) and \(a\) have the same sign. They don't need to convert to equations first. Students who score a 4 can do the algebra but get flustered when a problem requires graphical reasoning. The practice that builds this fluency: after doing a kinematic problem algebraically, sketch the x-t, v-t, and a-t graphs for the motion. Do this enough and graphs become a second language.
The question that reveals true understanding
Here's a question worth asking yourself: "A ball is thrown straight up. At the exact top of its flight, what is the ball's velocity, and what is its acceleration?" If you think \(v = 0\), \(a = 0\) — you're thinking about the ball as "at rest" at the peak and concluding nothing is happening. If you think \(v = 0\), \(a = -g\) — you understand that gravity doesn't care what the velocity is. This one question predicts whether you'll struggle with Unit 1 or fly through it.
Exam-day strategy
On kinematics FRQs, always do the following, in order: (1) draw a picture with coordinate axes labeled, (2) write down what you know and what you want with symbols (\(v_0, v, a, t, x_0, x\)), (3) identify which variable is missing from the problem and choose the equation that doesn't contain it, (4) solve symbolically, (5) plug in numbers last. AP graders give partial credit for setup even if the arithmetic fails at the end. Writing "\(v^2 = v_0^2 + 2a\Delta x\)" and identifying the right variables earns 2 of the 4 points on most problems, even if you don't finish.
On multiple choice: for projectile problems, eliminate any answer choice that has the horizontal velocity affecting time-of-flight, or the vertical velocity affecting horizontal range. Those are always wrong.
A surprising real-world connection
The NBA's three-point line is 7.24 m from the basket. A player's shot must leave their hand with just the right combination of angle and speed to land at the hoop, 3.05 m off the ground. A shooter's muscle memory is, without the player knowing it, a biological projectile-motion solver. When LeBron James adjusts his release angle a degree or two to compensate for a long-range shot, he's literally doing the calculation from problem 25 of this unit. The principles you're learning here govern every sport involving a ball — and every artillery targeting system, every rocket trajectory, every skateboarding ollie. You're not learning "physics problems." You're learning the grammar of how moving objects exist in the world.
One piece of advice before you move on
Don't memorize the three kinematic equations as a list. Memorize them as a family: each equation is missing exactly one of the five variables (\(v_0\), \(v\), \(a\), \(t\), \(\Delta x\)). When you see a problem, identify the missing variable — the one that's neither given nor asked for — and pick the equation without it. This turns equation selection from a guessing game into a reflex. It is, genuinely, the one habit that will make the rest of AP Physics 1 feel easier. Unit 2 will add forces. Unit 3 will add energy. But underneath everything, kinematics will still be the scaffolding. Build it solid here.