Motion in One Dimension

Key Definitions

Quantity	Type	Definition	Unit
Distance (d)	Scalar	Total path length travelled	m
Displacement (s or x)	Vector	Straight-line change in position from start to end	m
Speed (v)	Scalar	Distance ÷ time	m·s⁻¹
Velocity (v)	Vector	Displacement ÷ time	m·s⁻¹
Acceleration (a)	Vector	Change in velocity ÷ time (Δv/Δt)	m·s⁻²

Instantaneous vs Average: Average velocity = total displacement / total time. Instantaneous velocity is the velocity at a specific moment — it equals the slope of a position-time graph at that point.

Motion Graphs

Position-time (x-t) graph:

Slope = velocity
Horizontal line → object is stationary (v = 0)
Straight line (non-zero slope) → constant velocity
Curved line → changing velocity (acceleration)

Velocity-time (v-t) graph:

Slope = acceleration
Area under the graph = displacement
Horizontal line → constant velocity (a = 0)
Straight line with positive slope → uniform positive acceleration
Line crossing v = 0 → object reverses direction

Acceleration-time (a-t) graph:

Area under the graph = change in velocity (Δv)
Horizontal line → constant (uniform) acceleration

For uniform acceleration: v-t graph is a straight line. The area of the trapezium under v-t gives displacement.

Equations of Motion

These four equations apply only when acceleration is uniform (constant). Always define your positive direction first.

v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t

Symbol	Meaning	Unit
u	Initial velocity	m·s⁻¹
v	Final velocity	m·s⁻¹
a	Acceleration	m·s⁻²
s	Displacement	m
t	Time	s

Free Fall

Free fall is motion under gravity alone — no air resistance. The acceleration due to gravity near Earth's surface is g = 9.8 m·s⁻² downward.

Sign convention for free fall: Choose upward as positive → g = −9.8 m·s⁻² (or choose downward as positive → g = +9.8 m·s⁻²). Be consistent throughout your calculation.

Worked example: A ball is thrown upward at 14.7 m·s⁻¹. How high does it go?
Let up = +; u = +14.7 m·s⁻¹; a = −9.8 m·s⁻²; v = 0 at highest point.
v² = u² + 2as → 0 = 14.7² + 2(−9.8)s → s = 216.09/19.6 ≈ 11.0 m

At the highest point of a projectile thrown upward: v = 0 but a = g downward (≠ 0).

Ticker Tape Analysis

A ticker timer makes 50 dots per second (period = 0.02 s). The spacing between dots gives information about velocity and acceleration:

Equal spacing → constant velocity
Increasing spacing → acceleration (speeding up)
Decreasing spacing → deceleration (slowing down)

IEB Extension — Deriving Equations & Non-uniform Acceleration

Deriving from first principles: Starting from a = Δv/Δt = (v − u)/t, rearrange to get v = u + at. Substituting into s = average velocity × time: s = ½(u + v)t = ½(u + u + at)t = ut + ½at². Eliminate t between these two to get v² = u² + 2as.

Multi-stage problems: An object thrown upward from the edge of a cliff may have different conditions on the way up vs. the way down. Split the problem into stages, using the final velocity of stage 1 as the initial velocity of stage 2. Example: A ball thrown up at 10 m·s⁻¹ from a 15 m cliff. Find the time to hit the ground. (Take up as positive, displacement = −15 m at ground level.)

Non-uniform acceleration (qualitative): When acceleration varies with time, the equations of motion no longer apply. The instantaneous acceleration equals the slope of the v-t graph at each point. Displacement is still the area under the v-t graph (found by counting squares or integration in higher grades).

Show ALL working. Define your positive direction clearly for every free-fall problem. Use g = 9.8 m·s⁻² unless stated. Include neat, labelled sketches of motion graphs where requested.

Question 1

A car starts from rest and accelerates uniformly at 3 m·s⁻² for 8 seconds. (a) Calculate its final velocity. (b) Calculate the distance covered. (c) Sketch the v-t graph for this motion and shade the area that represents displacement.

Question 2

A v-t graph shows a straight line from v = 20 m·s⁻¹ at t = 0 to v = −4 m·s⁻¹ at t = 6 s. (a) Calculate the acceleration. (b) At what time does the object momentarily stop? (c) Calculate the total displacement from t = 0 to t = 6 s.

Question 3

A stone is dropped from a bridge 44.1 m above a river (take downward as positive, u = 0). Calculate: (a) the time to hit the water, (b) the velocity just before impact.

Question 4

A ball is thrown vertically upward at 19.6 m·s⁻¹ from ground level. Take upward as positive. Calculate: (a) the maximum height reached, (b) the time to reach maximum height, (c) the total time in the air before returning to the same level.

Question 5

A car travelling at 25 m·s⁻¹ brakes and decelerates at 5 m·s⁻² until it stops. (a) How long does braking take? (b) How far does the car travel while braking? (c) Sketch a neat x-t graph for the entire braking motion.

Question 6 — Position-Time Data Table

A trolley is released from rest at the top of a ramp. Its position is recorded every second:

Time (s)	0	1	2	3	4
Displacement (m)	0	2	8	18	32

(a) Calculate the trolley's average velocity between t = 1 s and t = 2 s. Show your working.
(b) Calculate the average velocity between t = 3 s and t = 4 s.
(c) Treating your answer to (a) as the instantaneous velocity at the midpoint t = 1.5 s, and your answer to (b) as the instantaneous velocity at t = 3.5 s, calculate the trolley's acceleration using a = Δv/Δt.
(d) The trolley starts from rest (u = 0). Using your acceleration from (c) in the equation s = ut + ½at², check whether it correctly predicts the displacement recorded at t = 4 s. Show your working and comment on the agreement.

Key Definitions

Motion Graphs

Equations of Motion

Free Fall

Ticker Tape Analysis

Motion Simulator — Ball on Track