Position, velocity, and acceleration are three linked ways to describe motion. - Position x(t): where an object is along a chosen axis at time t (can be shown in a motion diagram, equation, or x vs t graph). - Velocity vx(t): the rate of change of position—instantaneous velocity is dx/dt and equals the slope of the tangent on an x vs t graph. For constant acceleration use vx = vx0 + a t. Velocity tells you speed and direction (signed). - Acceleration ax(t): the rate of change of velocity—instantaneous acceleration is dvx/dt and equals the slope of the tangent on a v vs t graph. For constant acceleration a, use the three kinematic equations in the CED (vx = vx0 + a t; x = x0 + vx0 t + ½ a t^2; vx^2 = vx0^2 + 2a(x – x0)). Useful graph facts: area under v(t) = displacement (Δx = ∫ v dt); area under a(t) = change in velocity (Δv = ∫ a dt). For more review and AP-aligned practice, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16), the Unit 1 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-1), and 1000+ practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Start by listing what you know (v0, v, x0, x, a, t) and what you need. Check the CED condition: the three kinematic equations apply only for constant (uniform) acceleration: - v = v0 + a t - x = x0 + v0 t + ½ a t² - v² = v0² + 2 a (x − x0) Pick the equation that doesn’t introduce any unknowns you don’t have. Quick rules of thumb: - If time t is known (or asked for) and you need v → use v = v0 + a t. - If displacement x (or x − x0) and a are known but t isn’t → use v² = v0² + 2 a Δx. - If you need displacement and you know t → use x = x0 + v0 t + ½ a t². If acceleration isn’t constant, don’t use these—use graphs/derivatives or integrals (v = dx/dt, Δx = ∫ v dt, Δv = ∫ a dt). For free-fall near Earth use g ≈ 10 m/s² on the AP exam. For extra practice and quick reminders, check the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and the unit practice set (https://library.fiveable.me/practice/ap-physics-c-mechanics).

When you say “the slope of a position vs. time graph is the velocity,” you mean the rate at which position changes with time: v_x = dx/dt (CED 1.3.A.4.i). Concretely, at any point the instantaneous velocity equals the slope of the tangent line to the x(t) curve there. If the position graph is a straight line, the slope is constant → constant velocity; a steeper slope means a larger speed. A positive slope means motion in the +x direction, negative slope means motion in the −x direction. If the slope is changing (the position curve is curved), velocity is changing → there’s acceleration (CED 1.3.A.4.ii). You use this on the exam to translate between representations (graphs ↔ equations) and to find displacement (area under v(t)) per Topic 1.3. For a quick refresher, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Think of a motion diagram as a snapshot sequence that shows position, velocity, and acceleration at equal time steps. How to draw one (step-by-step): 1. Pick equal time intervals Δt and draw a dot for the object’s position at each time (more dots = longer time covered). 2. Spacing of dots shows speed: equal spacing → constant speed; increasing spacing → speeding up; decreasing → slowing down. 3. At each dot add a short arrow for velocity (vx) tangent to motion; length ∝ speed and direction shows sign. 4. Show acceleration (ax) with separate arrows: if acceleration is constant, the velocity arrows change by the same vector each Δt (length increases or decreases uniformly). For constant ax, v = v0 + a t and arrow-length changes reflect that. 5. Label times (t0, t1, …), and note signs (+/−). If it’s free fall use ag ≈ 10 m/s^2 downward. 6. Use the diagram to check graphs: position slope = velocity, velocity slope = acceleration (CED 1.3.A.4). For practice and AP-style translation between representations, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics). These diagrams often appear on Translation Between Representations FRQs, so practice converting between dots/arrows and x(t), v(t), a(t).

Because near Earth’s surface the dominant force on free objects is gravity from Earth, they accelerate toward Earth’s center. Newton’s law gives g = GM/r^2 (G and M are constants, r ≈ Earth’s radius), which evaluates to about 9.81 m/s^2. For AP problems we round that to g ≈ 10 m/s^2 (the CED explicitly allows this), so you’ll see “10 m/s^2 downward” used in kinematic equations like vx = vx0 + ax t and x = x0 + vx0 t + ½ a t^2. “Downward” just means the acceleration vector points toward Earth’s center; the sign in your equations depends on your chosen coordinate (take down as negative or positive consistently). g is effectively constant for most classroom problems because the change in r is tiny compared to Earth’s radius; it does vary slightly with altitude and latitude, but those differences aren’t tested on the AP. For a quick review, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and more practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Displacement from a v(t) graph = the area under the velocity curve between t1 and t2 (CED 1.3.A.4.iii). Mathematically Δx = ∫t1^t2 v(t) dt. Practically: - If v is constant: area = v·Δt (rectangle). Example: v = 3 m/s for 4 s → Δx = 12 m. - If v changes linearly: split into simple shapes (triangles/trapezoids). Triangle area = 1/2·base·height; trapezoid = (sum of parallel sides/2)·base. - If v goes negative, area below the time-axis counts negative (returns toward negative x). - For smooth curves, approximate by many small rectangles (Riemann sum) or compute the definite integral exactly when you have v(t). This idea is tested on the AP exam (Topic 1.3, representing motion). For more examples and practice, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and hundreds of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Average velocity = total displacement divided by total time: v_avg = Δx/Δt. It’s a single number that tells you how far and in what direction the object moved per unit time over an interval (useful on the exam for kinematics problems that give start & end positions). Instantaneous velocity is the object’s velocity at one specific moment. Mathematically v(t) = dx/dt and graphically it’s the slope of the tangent line to an x vs. t graph at that time. For constant acceleration you can use v = v0 + a t to get instantaneous v at a given t. Quick tips for AP Physics C: on position–time graphs find instantaneous v with a tangent (CED 1.3.A.4.i), use Δx = ∫ v dt for displacement (CED 1.3.A.4.iii), and remember average vs. instantaneous answers often show up in Translation Between Representations questions. For a focused review, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and more practice problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Read acceleration off a v vs t graph by finding the slope. Instantaneous acceleration ax(t) = dv/dt, so: - If the v(t) curve is a straight line, the acceleration is constant and equal to that line’s slope: a = (change in v)/(change in t). Example: v goes 2 → 8 m/s in 3 s → a = (8−2)/3 = 2 m/s^2. - If the curve isn’t a straight line, draw a tangent at the time you want and find its slope (rise/run) to get the instantaneous a. - Sign matters: positive slope → acceleration in the positive direction; negative slope → acceleration opposite the positive direction. - Also remember area under an a vs t curve gives Δv, and area under v vs t gives displacement (CED 1.3.A.4). For more practice and AP-aligned review, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and lots of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Think of velocity as how fast position changes: v_x = dx/dt (CED 1.3.A.4.i). Rearranging gives dx = v_x(t) dt. If you add up all those tiny position changes from t1 to t2 you get total displacement: Δx = ∫_{t1}^{t2} v_x(t) dt (CED 1.3.A.4.iii). That integral is exactly the “area under the velocity–time curve.” Intuition: for constant velocity the graph is a rectangle (v·Δt)—easy to see that area equals displacement. For non-constant v, slice the time interval into lots of tiny pieces where v is almost constant; each slice’s area (v·Δt) is a tiny displacement. Summing all slices and taking the limit gives the integral. On the exam you can use geometry for simple shapes or the integral/kinematic equations for general motion (see 1.3.A.2 and 1.3.A.4). For practice, try problems on the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and more practice at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

If a = 0 (no acceleration, constant velocity), the constant-acceleration kinematic equations simplify: - v = v0 + a t → v = v0 (velocity is constant). - x = x0 + v0 t + ½ a t^2 → x = x0 + v0 t (position changes linearly with time). - v^2 = v0^2 + 2a(x − x0) → v^2 = v0^2 (gives no new info about x when a = 0). Graphically: x vs. t is a straight line (slope = v0), v vs. t is a horizontal line at v0, and a vs. t is the horizontal axis at zero. Using calculus/area language from the CED: Δx = ∫ v(t) dt = v0 t and Δv = ∫ a(t) dt = 0. These are exactly the uniform (zero-acceleration) motion relations you should use on AP problems in Topic 1.3. For a quick refresher, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and more practice at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Quick rule: speeding up vs slowing down depends on velocity and acceleration directions. - From a velocity–time graph: look at the slope (the acceleration). If velocity and acceleration have the same sign, the object speeds up; if they have opposite signs, it slows down. Example: v = +3 m/s and a = +2 m/s^2 → speeding up; v = +3 m/s and a = −2 m/s^2 → slowing down. (Use ax = dv/dt from the CED.) - From a position–time graph: the slope is velocity. If the slope is getting steeper in the same direction (more positive or more negative) the object is speeding up; if the slope flattens toward zero the object is slowing down. Instantaneous velocity = tangent slope (CED 1.3.A.4.i). - From an acceleration–time graph: the area under a(t) gives Δv (CED 1.3.A.4.iv). If the area adds velocity in the same direction as current v, speed increases; if it subtracts, speed decreases. These are core AP CED ideas (use v = dx/dt and a = dv/dt). For a quick refresher and examples, check the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and try problems from the unit practice set (https://library.fiveable.me/practice/ap-physics-c-mechanics).

We write velocity as dx/dt because velocity is the instantaneous rate of change of position with time. dx/dt is the calculus way to say “how fast x changes at this exact moment”—geometrically it’s the slope of the tangent line on an x vs. t graph (CED 1.3.A.4.i). For constant acceleration you can use the kinematic formulas, but those are special-case algebraic results; dx/dt lets you handle any motion (non-uniform speed) and leads naturally to ax = dv/dt and the area/Integral relationships (Δx = ∫ v dt, Δv = ∫ a dt) in the CED (1.3.A.4.iii–iv). On the AP exam you’ll be expected to translate between graphs, equations, and tangents/areas (Topic 1.3). For a clear review, see the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and use the AP practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics) to get comfortable with slopes and integrals.

Distance is a scalar: it’s the total path length an object travels (always ≥ 0). Displacement is a vector: it’s the change in position from start to finish, Δx = xfinal − xinitial, with sign and direction. For straight-line motion displacement can be smaller than (or equal to) distance if the object reverses direction. On the AP C: Mechanics exam you must use displacement when relating position, velocity, and acceleration (v = dx/dt, Δx = ∫ v dt) and in kinematic equations for constant acceleration (e.g., x = x0 + v0 t + ½ a t²). Graphically, area under a v–t curve gives displacement, not distance—if the velocity changes sign you’d take separate positive areas to get total distance. If you want quick practice, review Topic 1.3’s study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and try problems from the Unit 1 page (https://library.fiveable.me/ap-physics-c-mechanics/unit-1) or the practice set (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Position, velocity, and acceleration graphs are just three ways of showing the same motion—they’re related by slopes (derivatives) and areas (integrals). - Velocity = slope of x(t). So where x(t) is flat, v = 0; where x(t) is steep positive, v > 0. Instantaneous v = dx/dt (CED 1.3.A.4.i). - Acceleration = slope of v(t). So constant a shows up as a straight line in v(t); instantaneous a = dv/dt (CED 1.3.A.4.ii). - Displacement = area under v(t) between t1 and t2: Δx = ∫ v dt (CED 1.3.A.4.iii). - Change in velocity = area under a(t): Δv = ∫ a dt (CED 1.3.A.4.iv). Quick examples: constant acceleration (a = constant) → x(t) is a parabola x = x0 + v0 t + ½ a t^2, v(t) is a line v = v0 + a t, a(t) is a horizontal line. For free fall use g ≈ 10 m/s^2. These graphs and translations are exactly the Translation Between Representations skills tested on the exam (see Topic 1.3 study guide: https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16). For extra practice, hundreds of problems are at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Treat projectile motion as two independent 1-D motions: horizontal (x) and vertical (y). Use the constant-acceleration kinematic equations separately for each axis from the CED. Steps: 1. Split the initial velocity: v_x0 = v0 cosθ, v_y0 = v0 sinθ. 2. Use a_x = 0 and a_y = −g (use g ≈ 10 m/s^2 on AP problems). Apply: - x = x0 + v_x0 t (or x = x0 + v_x0 t + ½ a_x t^2 but a_x=0) - y = y0 + v_y0 t − ½ g t^2 - v_y = v_y0 − g t (or v_y^2 = v_y0^2 − 2g(y−y0) when you need height/time without t) 3. Solve the vertical equation for time(s) t of interest (e.g., when y = 0 for range), then plug that t into the horizontal equation to get range or x-position. 4. For max height set v_y = 0 and use v_y^2 relation or solve t = v_y0/g and plug into y. 5. Check signs and units; sketch motion/velocity-time graphs to verify (CED 1.3.A.4). Want worked examples and timed practice? See the Topic 1.3 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16) and hundreds of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).