AP^® · Algebra-Based

AP^® Physics 1

All eight College Board units, the three science practices, and the equation sheet: built for the 42 MC and the 4 FRQs.

Start Unit 1 free. Unit 1: Kinematics is open to everyone, no account needed. Other topics are locked.

Unit 1: Kinematics

THE BIG PICTURE. Unit 1 (10–15% of the MC) is the language layer for everything downstream. Forces (Unit 2), energy (Unit 3), momentum (Unit 4), and the rotational analogues (Units 5 & 6) are all about motion, so if your kinematics is sloppy, every later unit inherits that sloppiness. Two ideas dominate: (1) vectors vs. scalars, and (2) the three constant-acceleration equations plus their graphical interpretation.

SCALARS, VECTORS, AND THE THREE KINEMATIC QUANTITIES

A scalar has magnitude only (mass, speed, distance, time, energy). A vector has magnitude AND direction (displacement, velocity, acceleration, force, momentum). In 1D, vector direction is encoded by a sign (+/−); the sign convention is your choice but must be applied consistently across the whole problem.

Displacement $\Delta x = x_f - x_i$ (vector). Distance is the scalar path length and is $\geq |\Delta x|$ .
Velocity $\vec v$ (vector) = rate of change of position. Average $\bar v = \Delta x / \Delta t$ . Instantaneous $v$ = slope of the tangent line on the $x$ vs $t$ graph. Speed is $|\vec v|$ .
Acceleration $\vec a$ (vector) = rate of change of velocity. Average $\bar a = \Delta v / \Delta t$ . Instantaneous $a$ = slope of the tangent line on the $v$ vs $t$ graph.

Critical sign convention: acceleration is NOT "speeding up positive, slowing down negative." Acceleration sign tracks the vector direction. An object can have negative acceleration while speeding up (moving in the $-x$ direction faster). The rule: same sign on $v$ and $a$ means speeding up; opposite signs mean slowing down.

PRACTICE: SPEEDING UP OR SLOWING DOWN?

Take $+x$ to the right. Cover the last column and predict it from the signs alone.

Sign of $v$	Sign of $a$	What the object is doing
+	+	moving right, speeding up
+	−	moving right, slowing down
−	−	moving left, speeding up (negative acceleration, yet faster)
−	+	moving left, slowing down
0 (for an instant)	nonzero	momentarily at rest; it turns around if $v$ changes sign
any	0	constant velocity (same speed, same direction)

THE THREE CONSTANT-ACCELERATION EQUATIONS (memorize; on the equation sheet)

For constant $a$ : $v = v_0 + at$ $x = x_0 + v_0 t + \tfrac{1}{2} a t^2$ $v^2 = v_0^2 + 2a(x - x_0)$

A fourth useful relation (not on the sheet, but derivable): $\bar v = \tfrac{1}{2}(v_0 + v)$ for constant $a$ .

Choosing which equation: list your knowns. Whichever equation contains every known plus the one unknown is the equation. The third equation ( $v^2 = v_0^2 + 2a\Delta x$ ) is the only one without $t$ : use it when time is neither given nor asked.

FREE FALL

( $g = 9.8$ m/s $^2$ near Earth's surface, given on the equation sheet alongside $G$ ). Near Earth's surface, $g = 9.8$ m/s $^2$ downward (call this $-y$ if up is positive). Free-fall problems are constant-acceleration problems with $a = -g$ . Key consequences:

An object thrown straight up reaches max height when $v = 0$ , then returns with the same speed (in opposite direction) it was launched.
Time up = time down for a symmetric trajectory.
At any height on the way down, speed equals speed at the same height on the way up.

Which value of g? The equation sheet lists $g = 9.8$ m/s $^2$ , and the CED adds that exam questions needing a number for $g$ use $10$ m/s $^2$ (correct use of $9.8$ is not penalized). This overview uses $9.8$ ; the practice FRQs in this guide use $10$ . Follow whatever the question states.

CASE STUDY: THE HAMMER AND THE FEATHER ON THE MOON (1971)

Near the end of the last Apollo 15 moonwalk on August 2, 1971, Commander David Scott held up a geology hammer (about 1.32 kg) and a falcon feather (about 0.03 kg) and released them together from roughly 1.6 m. With essentially no air on the Moon, they landed at the same moment on live television. Taking the Moon's $g \approx 1.6$ m/s $^2$ (about one-sixth of Earth's), $\Delta y = \tfrac{1}{2}gt^2$ gives a fall time of about $t = \sqrt{2(1.6)/1.6} \approx 1.4$ s, versus about $0.57$ s for the same drop in a vacuum on Earth.

Concept: free fall. With gravity the only force, every object near the same body has the same acceleration, whatever its mass.
Representation: two identical position-time graphs (same parabola) for hammer and feather; a $v$ - $t$ graph that is one straight line with slope $-g$ .
Lesson: on Earth the feather lags because of air resistance, not because gravity treats it differently. AP problems assume negligible air resistance unless they say otherwise.

MOTION GRAPHS

The three graphs $x(t)$ , $v(t)$ , $a(t)$ are linked by derivatives (slopes) and integrals (areas):

Graph	Slope gives	Area under curve gives
$x$ vs $t$	velocity $v$	(no physical meaning)
$v$ vs $t$	acceleration $a$	displacement $\Delta x$
$a$ vs $t$	jerk (out of scope)	change in velocity $\Delta v$

The single highest-yield MC skill in Unit 1 is reading these graphs fluently: identifying when the object is moving forward/backward, speeding up/slowing down, at rest, or changing direction. Object turns around when $v = 0$ AND $a \neq 0$ (the $x$ -graph hits a local max or min).

Three stacked graphs for one cart trip: position vs time curving up, straight, then curving down to 20 m; velocity vs time rising to 4 m/s, flat, then falling to zero with the trapezoid area shaded as 20 m; acceleration vs time at plus 2, zero, then minus 2 m/s squared with the areas shaded as plus and minus 4 m/s. — **One trip, three graphs** Each graph is the slope of the one above it, and the area under each graph is the change in the one above it. The velocity graph's area gives the 20 m displacement; the acceleration graph's areas give the +4 m/s and -4 m/s velocity changes.

WORKED EXAMPLE: ONE TRIP, THREE GRAPHS

A cart starts from rest, speeds up at $2$ m/s $^2$ for $2$ s, cruises at $4$ m/s for $3$ s, then brakes at $2$ m/s $^2$ until it stops at $t = 7$ s (see the figure). Find (a) the displacement, (b) the average velocity, (c) whether the cart is speeding up or slowing down at $t = 6$ s, and (d) its position at $t = 5$ s.

(a) Area under $v$ - $t$ . Triangle + rectangle + triangle: $\tfrac{1}{2}(2)(4) + (3)(4) + \tfrac{1}{2}(2)(4) = 4 + 12 + 4 = 20$ m.
(b) Average velocity. $\bar v = \Delta x/\Delta t = 20/7 \approx 2.9$ m/s. Not the average of the three speeds: it weights each by its time.
(c) Signs at $t = 6$ s. $v = +2$ m/s and $a = -2$ m/s $^2$ : opposite signs, so the cart is slowing down (the $x$ - $t$ graph is concave down there).
(d) Position at $t = 5$ s. Area from $0$ to $5$ s: $4 + 12 = 16$ m, matching the $x$ - $t$ graph.

A Translation Between Representations question might give you only the $v$ - $t$ graph and ask you to sketch the other two. Check your sketch with the rules in the figure: where $v$ is constant, $x$ is a straight line and $a = 0$ ; where $v$ changes, $x$ curves.

CASE STUDY: 9.58 SECONDS IN BERLIN (2009)

On August 16, 2009, Usain Bolt ran 100 m in 9.58 s, which is still the world record. Researchers published his split times at every 10 m:

Distance (m)	10	20	30	40	50	60	70	80	90	100
Time (s)	1.90	2.88	3.80	4.63	5.46	6.29	7.11	7.92	8.75	9.58

Concept: average vs instantaneous velocity. His average velocity was $100/9.58 \approx 10.4$ m/s, but from 60 m to 80 m he covered $20$ m in $1.63$ s, about $12.3$ m/s.
Representation: plotted as position vs time, the first few points curve upward (acceleration), then the graph becomes almost straight (nearly constant top speed). The first 10 m, which includes reacting to the gun, took almost 2 s.
Lesson: a single average hides the shape of the motion. Slopes of short intervals approximate the instantaneous velocity.

REFERENCE FRAMES AND RELATIVE MOTION

The velocity of object A relative to object B is $\vec v_{AB} = \vec v_A - \vec v_B$ . In AP Physics 1 this is restricted to one dimension, where it is just signed addition or subtraction. Classic setup: a boat moving with velocity $\vec v_{BW}$ relative to the water, along a river whose current is $\vec v_{WG}$ relative to the ground. The boat's velocity relative to ground is $\vec v_{BG} = \vec v_{BW} + \vec v_{WG}$ . The order of subscripts matters: $\vec v_{AB} = -\vec v_{BA}$ .

WORKED EXAMPLE: UPSTREAM AND BACK

A boat moves at $5.0$ m/s relative to the water in a river whose current is $2.0$ m/s. It travels $600$ m upstream and returns. How long does the round trip take, and how does that compare with still water?

Step 1, relative velocities (1D). Upstream: $5.0 - 2.0 = 3.0$ m/s relative to the ground. Downstream: $5.0 + 2.0 = 7.0$ m/s.
Step 2, times. Up: $600/3.0 = 200$ s. Down: $600/7.0 \approx 86$ s. Total $\approx 286$ s.
Step 3, compare. In still water: $1200/5.0 = 240$ s. The current costs more time upstream than it saves downstream, because the boat spends longer going slowly.

PROJECTILE MOTION (THE HEADLINE 2D APPLICATION)

A projectile experiences only gravity after launch. Decompose:

Horizontal: $a_x = 0$ , so $v_x = v_{0x}$ constant and $x = v_{0x} t$ .
Vertical: $a_y = -g$ , so $v_y = v_{0y} - gt$ and $y = v_{0y} t - \tfrac{1}{2} g t^2$ .

The two axes are decoupled: time is the only shared variable. Solve the vertical equations first (they typically give you $t$ via the height condition), then plug $t$ into the horizontal equation for range.

Launch angle $\theta$ : $v_{0x} = v_0 \cos\theta$ and $v_{0y} = v_0 \sin\theta$ . For launch and landing at the same height on level ground, range $R = (v_0^2 \sin 2\theta) / g$ (NOT on the equation sheet but derivable). Maximum range is at $\theta = 45°$ ; complementary angles ( $30°$ and $60°$ ) give the same range.

Common trap: at the peak of a projectile's arc, $v_y = 0$ but $v_x \neq 0$ , so the speed is non-zero and acceleration is still $-g$ (gravity does not turn off).

Parabolic path of a ball launched at 20 m/s and 37 degrees, with horizontal velocity arrows of equal length at five times and vertical velocity arrows shrinking from +12 m/s to zero at the top and growing to -12 m/s at landing; maximum height 7.35 m and range 39.2 m marked. — **Components of projectile motion** The horizontal velocity stays at 16 m/s the whole flight, while the vertical velocity drops by 9.8 m/s every second. At the top only the horizontal part remains, and the acceleration is still g downward.

WORKED EXAMPLE: A 37° LAUNCH FROM LEVEL GROUND

A ball is launched at $20$ m/s at $37°$ above the horizontal and lands at launch height. Find the time to the top, the maximum height, the time of flight, and the range. (The equation sheet gives $\sin 37° = 0.6$ and $\cos 37° = 0.8$ .)

Components. $v_{0x} = 20(0.8) = 16$ m/s and $v_{0y} = 20(0.6) = 12$ m/s.
Top. $v_y = 0$ : $0 = 12 - 9.8t$ , so $t = 1.22$ s. Height: $v_y^2 = v_{0y}^2 - 2g\Delta y$ gives $\Delta y = 12^2/(2 \cdot 9.8) = 7.35$ m.
Flight time. Symmetric trajectory: $2(1.22) = 2.45$ s.
Range. Horizontal motion is uniform: $R = v_x t = 16(2.45) = 39.2$ m. Check with the derived formula: $v_0^2 \sin 2\theta/g = 400 \sin 74°/9.8 \approx 39.2$ m.
Speed at the top. Only $v_x$ remains: $16$ m/s, not zero.

EXAM CONNECTIONS. Unit 1 content is a natural fit for the Translation Between Representations FRQ (Q2). Expect to be handed a position-time or velocity-time graph and asked to sketch the others, identify intervals of constant velocity vs. acceleration, or compute average vs. instantaneous quantities. MC questions reward fluent sign-convention reasoning and the ability to pick the right kinematic equation in one read.

Key Terms

Scalar

Quantity with magnitude only (mass, time, speed, distance, energy). Sign carries no directional meaning.

Vector

Quantity with magnitude AND direction (displacement, velocity, acceleration, force, momentum). In 1D, encoded as signed value; in 2D, as components.

Displacement ( $\Delta x$ )

Vector change in position: $\Delta x = x_f - x_i$ . Independent of path. Magnitude $\leq$ distance travelled.

Velocity

Vector rate of change of position. Average $\bar v = \Delta x / \Delta t$ ; instantaneous $v$ = slope of the tangent line on the $x$ vs $t$ graph.

Acceleration

Vector rate of change of velocity. Instantaneous $a$ = slope of the tangent line on the $v$ vs $t$ graph. Sign tracks vector direction, NOT speeding up vs slowing down. Same-sign $v$ and $a$ → speeding up; opposite signs → slowing down.

Constant-Acceleration Kinematic Equations

$v = v_0 + at$ ; $x = x_0 + v_0 t + \tfrac{1}{2} a t^2$ ; $v^2 = v_0^2 + 2a\Delta x$ . The third has no $t$ : use when time is neither given nor asked.

Free Fall

Motion under gravity alone. Near Earth, $a = g = 9.8$ m/s $^2$ downward. Air resistance ignored unless stated.

Motion Graph Slopes and Areas

Slope of $x$ vs $t$ = $v$ ; slope of $v$ vs $t$ = $a$ . Area under $v$ vs $t$ = $\Delta x$ ; area under $a$ vs $t$ = $\Delta v$ .

Reference Frame

A coordinate system in which motion is described. Velocities transform via $\vec v_{AC} = \vec v_{AB} + \vec v_{BC}$ .

Relative Velocity

Velocity of A as measured in B's frame: $\vec v_{AB} = \vec v_A - \vec v_B$ . Note $\vec v_{AB} = -\vec v_{BA}$ .

Projectile

Object in free flight under gravity alone. Horizontal motion ( $a_x = 0$ ) and vertical motion ( $a_y = -g$ ) decouple; time is the shared variable.

Range Equation

For launch and landing at the same height: $R = (v_0^2 \sin 2\theta) / g$ . Max at $\theta = 45°$ ; complementary angles give equal ranges.

Inertial Reference Frame

A frame in which Newton's first law holds (no fictitious forces). All AP Physics 1 problems assume an inertial frame unless stated.

Exam Tips

Always pick a positive direction before starting any kinematics problem and stick to it: confusing sign conventions costs more points than any other Unit 1 error.
When choosing a kinematic equation, list your knowns and unknowns first. Only one of the three constant- $a$ equations contains all your knowns plus the one unknown.
For projectile problems, decouple horizontal and vertical immediately: they share only $t$ . Solve the vertical equations first (they give $t$ from the height condition), then plug into horizontal.
Remember that at the peak of a projectile trajectory, $a = g$ downward, not zero. The vertical velocity is zero, but acceleration is unchanged.
Read motion graphs by zone: identify intervals of constant $v$ (flat slope on $v$ - $t$ ), constant $a$ (straight line on $v$ - $t$ ), or zero motion. The Translation Between Representations FRQ rewards this segment-by-segment reading.
For the AP equation sheet, constant acceleration is the only kinematics formula given: variable-acceleration problems require you to use slopes and areas of motion graphs, not formulas.

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AP® Physics 1

Unit 1: Kinematics

SCALARS, VECTORS, AND THE THREE KINEMATIC QUANTITIES

PRACTICE: SPEEDING UP OR SLOWING DOWN?

THE THREE CONSTANT-ACCELERATION EQUATIONS (memorize; on the equation sheet)

FREE FALL

CASE STUDY: THE HAMMER AND THE FEATHER ON THE MOON (1971)

MOTION GRAPHS

WORKED EXAMPLE: ONE TRIP, THREE GRAPHS

CASE STUDY: 9.58 SECONDS IN BERLIN (2009)

REFERENCE FRAMES AND RELATIVE MOTION

WORKED EXAMPLE: UPSTREAM AND BACK

PROJECTILE MOTION (THE HEADLINE 2D APPLICATION)