AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/1/1-3)
Unit 1 · Topic 1.3
1.3 Representing Motion
The same motion can be shown as a motion diagram, a graph, an equation or a sentence, and AP questions constantly ask you to move between them. Slopes and areas link the position, velocity and acceleration graphs, and the three kinematic equations solve any problem with constant acceleration, including free fall.
Key terms
- motion diagram
- position–time graph
- velocity–time graph
- slope and area under a graph
- kinematic equations
- free fall
Motion diagrams
A motion diagram shows an object's position as a dot at equal time intervals, like a strobe photo. Evenly spaced dots mean constant velocity. Spreading dots mean speeding up, and bunching dots mean slowing down. You can add velocity arrows at each dot and one acceleration arrow for the whole diagram.
Reading motion graphs
Position–time (x–t): the slope is velocity. A straight line means constant velocity, a steeper line means faster motion, and a flat line means the object is at rest. A curve means the velocity is changing; its instantaneous velocity at any moment is the slope of the tangent line there.
Velocity–time (v–t): the slope is acceleration, and the area between the curve and the time axis is displacement. Area below the axis counts as negative displacement. Where the line crosses v = 0, the object momentarily stops and may turn around.
Acceleration–time (a–t): the area under the curve is the change in velocity, Δv.
On an x–t graph, curvature shows acceleration. A curve that bends upward (concave up, like a bowl) means the slope is increasing, so the acceleration is positive. A curve that bends downward means negative acceleration. Constant acceleration makes the x–t graph a parabola and the v–t graph a straight line.
Read the axis label before anything else. The same straight, rising line means "moving at constant velocity" on an x–t graph but "speeding up steadily" on a v–t graph.
The kinematic equations
When acceleration is constant, three equations describe motion along one axis. They appear on the equation sheet:
- v = v₀ + at (no Δx in it)
- x = x₀ + v₀t + ½at² (no final velocity v in it)
- v² = v₀² + 2a(x − x₀) (no time t in it)
| If the problem doesn't give or ask for… | use |
|---|---|
| displacement | v = v₀ + at |
| final velocity | x = x₀ + v₀t + ½at² |
| time | v² = v₀² + 2a(x − x₀) |
Free fall
An object in free fall is acted on only by gravity, which is the default assumption when air resistance is negligible. Near Earth's surface its acceleration is about 9.8 m/s² straight down, no matter its mass and no matter whether it's moving up, down or sitting at the top of its path. The course framework says exam questions that need a number use g = 10 m/s², but 9.8 or 9.81 m/s² is never marked wrong, so use whichever value the problem gives.
With up as positive, a = −g. At the highest point the velocity is zero for an instant, but the acceleration is still 9.8 m/s² downward. If the object lands at the height it was thrown from, the trip up and the trip down take equal times and it lands at the same speed it was thrown.
For a ball thrown straight up and caught at the same height, the three graphs look like this (up positive): the y–t graph is an upside-down parabola peaking at the top of the flight; the v–t graph is a straight line with slope −g, starting positive and crossing zero at the top; and the a–t graph is a flat line at −g the whole time.
When acceleration isn't constant
The kinematic equations aren't magic. With constant acceleration, the v–t graph is a straight line, and the area under it from 0 to t is a rectangle (v₀t) plus a triangle (½at²). That's exactly the displacement equation.
The kinematic equations fail when acceleration changes. On the exam you only treat changing acceleration qualitatively: sketching or interpreting graphs. For example, a skydiver with air resistance has a v–t graph that starts steep and then levels off toward a top speed, because the acceleration shrinks toward zero.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Speeding up from rest
A car starts from rest and accelerates at a constant 3.0 m/s² for 4.0 s. How far does it travel, and how fast is it going at the end?
Show the solutionHide the solution
- Step 1: Knowns: v₀ = 0, a = 3.0 m/s², t = 4.0 s. Unknowns: displacement and v.
- Step 2: Displacement: x − x₀ = v₀t + ½at² = 0 + ½(3.0)(4.0)² = 24 m.
- Step 3: Final velocity: v = v₀ + at = 0 + (3.0)(4.0) = 12 m/s.
- Step 4: Check with the third equation: v² = 0 + 2(3.0)(24) = 144, so v = 12 m/s. ✓
Answer: 24 m; 12 m/s
- Example 2Calculator allowed
Ball thrown upward from above the ground
You throw a ball straight up at 15 m/s from a point 2.0 m above the ground. Use g = 9.8 m/s² and ignore air resistance. Find (a) how high above the ground it rises, (b) how long it takes to hit the ground, and (c) its acceleration at the top of its path.
Show the solutionHide the solution
- Step 1: Take up as positive and the launch point as y₀ = 0, so a = −9.8 m/s².
- Step 2: (a) At the top v = 0. Use v² = v₀² + 2a(y − y₀): 0 = 15² + 2(−9.8)(y), so y = 225 ÷ 19.6 ≈ 11.5 m above the launch point. Add the 2.0 m: about 13.5 m above the ground.
- Step 3: (b) The ground is at y = −2.0 m. Use y = v₀t + ½at²: −2.0 = 15t − 4.9t². Rearranged, 4.9t² − 15t − 2.0 = 0. The quadratic formula gives t ≈ 3.19 s or t ≈ −0.13 s. Time can't be negative here, so t ≈ 3.2 s.
- Step 4: (c) The ball is still in free fall at the top, so a = 9.8 m/s² downward. Its velocity is zero there, but its acceleration is not.
Answer: (a) about 13.5 m above the ground; (b) about 3.2 s; (c) 9.8 m/s² downward
- Example 3Calculator allowed
Displacement from a velocity–time graph (classic trap)
A cart's velocity changes steadily from +6.0 m/s at t = 0 to −2.0 m/s at t = 4.0 s, so its v–t graph is a straight line. Find (a) its acceleration, (b) its displacement from 0 to 4.0 s, and (c) the distance it travels.
Show the solutionHide the solution
- Step 1: (a) Slope = Δv/Δt = (−2.0 − 6.0) ÷ 4.0 = −2.0 m/s².
- Step 2: The line crosses v = 0 where 6.0 − 2.0t = 0, at t = 3.0 s. That's when the cart turns around.
- Step 3: (b) Area above the axis (0 to 3.0 s): ½(3.0 s)(6.0 m/s) = +9.0 m. Area below the axis (3.0 to 4.0 s): ½(1.0 s)(−2.0 m/s) = −1.0 m. Displacement = 9.0 − 1.0 = +8.0 m.
- Step 4: (c) Distance counts both pieces as positive: 9.0 + 1.0 = 10.0 m. The trap is treating the area below the axis as positive when you want displacement.
Answer: (a) −2.0 m/s²; (b) +8.0 m; (c) 10.0 m
Common mistakes
- Reading an x–t graph as if it were a v–t graph. Check the vertical axis first; the same shape means different motion on each.
- Saying the acceleration is zero at the top of a throw. Only the velocity is zero there; gravity still gives 9.8 m/s² downward.
- Using the kinematic equations when the acceleration changes. They only work for constant acceleration; otherwise reason with graphs.
- Giving "−9.8" to a free-falling object in an up-positive problem and then also adding a minus sign in the equation. Put the sign in the value of a once.
On the exam
- Graph translation is a favorite: given an x–t graph, sketch the v–t graph (or the other way around). Match slopes to values, and line up turning points and zero crossings at the same times.
- On free response, start from an equation on the sheet, substitute with units and show your setup. A derivation in symbols (for example, h = v₀²/2g) usually comes before any numbers.
- Questions with no numbers often ask how a quantity changes. If the launch speed doubles, the maximum height (v₀²/2g) quadruples.
Connected topics
Videos
Check yourself
4 questions on 1.3 Representing Motion. Pick an answer to see if you got it, and why.
| Time (s) | Velocity (m/s) |
|---|---|
| 0.0 | +6.0 |
| 1.0 | +4.0 |
| 2.0 | +2.0 |
| 3.0 | 0.0 |
| 4.0 | −2.0 |
Experimental data: a cart moves along a straight track, and its velocity is recorded every second. Its velocity–time graph is a straight line.
What is the cart's acceleration?
What is the cart's displacement from t = 0 to t = 4.0 s?
What total distance does the cart travel from t = 0 to t = 4.0 s?
At what time is the cart farthest in the positive direction from its starting point?
0 of 4 answered