3.0Accelerated Motion
Overview: Accelerated Motion: Acceleration, Constant-Acceleration Equations and Free Fall — American System Physics Revision Notes & Topical Questions
5 min read 7 parts Lesson 2 of 5 Unit quiz
Lesson summary
Define and calculate acceleration.In free fall, all objects accelerate equally (ignoring air resistance).At the top of a vertical throw, velocity is zero but acceleration is still g.Area under a v–t graph gives displacement.Deceleration is acceleration opposite to velocity.Air re…
- Take up as positive
- A car goes from 0 to 24 m/s in 6.0 s. Find a and distance
- A ball is dropped from 20 m. Speed on hitting the ground?
- Velocity–time graph
- Uniform acceleration
- Acceleration due to gravity (g)
Section 1 of 7Overview: Accelerated Motion: Acceleration, Constant-Acceleration Equations and Free Fall
Take up as positive
Take up as positive a = −9.8 m/s².
Time up equals time down for symmetric vertical motion.
v–t slope zero means constant velocity.
Units of acceleration are m/s².
Reaction time adds distance before braking begins.
Section 2 of 7Process & sequence
1
Choose a positive direction.
2
List vi, vf, a, t, Δx (know three).
3
Pick the equation that lacks the unknown you do not need.
4
Solve symbolically, then substitute.
5
Check sign and units.
Vertical throwwrite these steps in this order; each step is a marking point.
- Launched upward
- Slows at g
- v = 0 at top
- Falls, speeding up
- Returns with same speed
Section 3 of 7Compare
| Equation | Missing variable |
|---|---|
| vf = vi + at | displacement |
| xf = xi + vit + ½at² | final velocity |
| vf² = vi² + 2aΔx | time |
| Δx = ½(vi + vf)t | acceleration |
| Velocity | Acceleration | Result |
|---|---|---|
| + | + | Speeding up |
| + | − | Slowing down |
| − | − | Speeding up (backward) |
| − | + | Slowing down |
Section 4 of 7Worked examples
1A car goes from 0 to 24 m/s in 6.0 s. Find a and distance.
A car goes from 0 to 24 m/s in 6.0 s. Find a and distance.
- a = 4.0 m/s²; Δx = ½ × 4.0 × 36 = 72 m.
2A ball is dropped from 20 m. Speed on hitting the ground?
A ball is dropped from 20 m. Speed on hitting the ground?
- v = √(2 × 9.8 × 20) = 19.8 m/s.
Section 5 of 7Key terms
Section 6 of 7Command terms
Command terms
Identify / Describe (FRQ)
A short, specific statement; describing requires relevant characteristics, not a single word.
Explain / Justify (FRQ)
Give the cause-and-effect reasoning or evidence; a restated fact does not score.
Calculate
Show setup, substitution and the answer with units.
Predict
State the expected outcome; add a reason only if asked to justify.
MCQ technique
Read every option; eliminate distractors that are true statements but do not answer the question; watch for NOT/EXCEPT.
Section 7 of 7Exam technique
Keywords the mark scheme looks for in this lesson
Keywords the mark scheme looks for in this lesson
Acceleration · Velocity–time graph · Uniform acceleration · Free fall · Acceleration due to gravity (g)
Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.
How to get full marks (examiner guidance)
1
Answer exactly what is asked; extra correct material does not earn credit, and contradictory statements can cancel credit (AP FRQ).
2
Use precise vocabulary (e.g. “active site”, “concentration gradient”, “allele”), never “it/stuff/things”.
3
Show units and working in calculations.
4
In MCQs, predict the answer before reading the options, then eliminate.
Instant feedbackCheck yourself
Q1. Area under a v–t graph gives
- A. acceleration
- B. displacement
- C. force
- D. mass
Show answer
B. v × t.
Q2. A ball thrown up: acceleration at the top is
- A. 0
- B. 9.8 m/s² down
- C. 9.8 m/s² up
- D. maximum velocity
Show answer
B. Gravity still acts.
Q3. From rest, a = 2 m/s² for 5 s. Final speed
- A. 2 m/s
- B. 5 m/s
- C. 10 m/s
- D. 25 m/s
Show answer
C. v = at.
Q4. Which equation has no time variable?
1
A. v = vi + at
2
B. v² = vi² + 2aΔx
3
C. x = vit + ½at²
4
D. a = Δv/Δt
Show answer
B. Time-free.
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