Motion along a line
For an object moving along a straight line:
| Quantity | Symbol | Relationship |
|---|---|---|
| Position | s(t) | where it is |
| Velocity | v(t) | s′(t) |
| Acceleration | a(t) | v′(t) = s″(t) |
Velocity
- v > 0: moving forward (right or up)
- v < 0: moving backward (left or down)
- v = 0: at rest (momentarily stopped, possibly turning around)
Speed = |v|: how fast, regardless of direction.
Acceleration
- a > 0: velocity increasing
- a < 0: velocity decreasing
Speeding up or slowing down?
- Speeding up when v and a have the same sign.
- Slowing down when they have opposite signs.
Example
s(t) = t³ − 9t² + 24t (metres, seconds), t ≥ 0
- v(t) = 3t² − 18t + 24 = 3(t − 2)(t − 4)
- a(t) = 6t − 18
At rest: v = 0 → t = 2 s and t = 4 s
Direction:
| Interval | v | Direction |
|---|---|---|
| 0 < t < 2 | + | forward |
| 2 < t < 4 | − | backward |
| t > 4 | + | forward |
Positions: s(0) = 0, s(2) = 20, s(4) = 16
Total distance from 0 to 4 s: 20 (out) + 4 (back) = 24 m
Displacement (change in position) from 0 to 4 s: 16 − 0 = 16 m
At t = 1: v = 9, a = −12 → opposite signs → slowing down.
Vertical motion
Near Earth’s surface, ignoring air resistance:
h(t) = −4.9t² + v₀t + h₀
- v(t) = −9.8t + v₀
- a(t) = −9.8 m/s² (gravity)
Example: A ball is thrown up at 19.6 m/s from 2 m.
- Max height when v = 0: t = 2 s → h(2) = −19.6 + 39.2 + 2 = 21.6 m
- Lands when h = 0: t ≈ 4.1 s; velocity ≈ −20.6 m/s
Distance vs displacement
| Displacement | Total distance |
|---|---|
| final − initial position | sum of all distances travelled |
| can be negative | always positive |
| ignores turnarounds | counts each leg |
To find total distance, split the time at each turnaround (where v changes sign).
Worked examples
Example 1: s(t) = 5t² − 3t. Velocity at t = 2? v = 10t − 3 → 17.
Example 2: v(t) = 4 − t². When at rest (t ≥ 0)? t = 2.
Example 3: v = −3, a = −2. Speeding up or slowing down? Same sign → speeding up.
Common mistakes
- Confusing speed and velocity.
- Thinking negative acceleration always means slowing down.
- Using displacement when the question asks for total distance.
Try it yourself
- s(t) = t² − 6t + 5. When is the object at rest?
- Velocity and acceleration at t = 1 for question 1?
- Is the object speeding up or slowing down at t = 1?
- Total distance from t = 0 to t = 5 for question 1?
Answers: 1) t = 3 2) v = −4, a = 2 3) slowing down (opposite signs) 4) 9 + 4 = 13
Quick recap
- Velocity = s′(t); acceleration = s″(t).
- At rest when v = 0; speed = |v|.
- Displacement ≠ total distance when the object turns around.