Motion Along a Straight Line

Displacement, velocity, acceleration, motion graphs, and SUVAT equations

SPEC 3.4.1.3 • MECHANICS AND MATERIALS

Key Definitions

Scalar vs Vector

Distance is a scalar (magnitude only) — the total length of the path travelled.

Displacement is a vector (magnitude and direction) — the straight-line distance from start to finish.

Speed (Scalar)

Rate of change of distance

speed = \frac{distance}{time}

Velocity (Vector)

Rate of change of displacement

v = \frac{Δ s}{Δ t}

Acceleration

Definition

Acceleration is the rate of change of velocity.

a = \frac{Δ v}{Δ t} = \frac{v - u}{t}

Units: m s⁻² (metres per second squared)

Key Point

Acceleration is a vector. A negative acceleration means the object is either slowing down (if moving in positive direction) or speeding up in the negative direction.

Common Confusion

Deceleration is not a physics term — it's just negative acceleration. Always use "acceleration" with appropriate sign.

Average vs Instantaneous

Average Velocity

Total displacement divided by total time

\overset{v}{ˉ} = \frac{Δ s}{Δ t}

Gives the overall rate over a time interval

Instantaneous Velocity

Velocity at a specific moment in time

v = Δ t \to 0 lim \frac{Δ s}{Δ t}

Found from gradient of s-t graph at a point

Example

A car travels 100 km in 2 hours. Its average speed is 50 km/h. But its instantaneous speed varied — sometimes 80 km/h on the motorway, sometimes 0 km/h at traffic lights.

Worked Example01

A cyclist travels 240 m in 30 s, then stops for 10 s, then travels another 160 m in 20 s. Calculate: (a) the average speed for the whole journey, (b) the average velocity if the cyclist ends up 400 m from the start in a straight line.

1

Total distance = 240 + 0 + 160 = 400 m

2

Total time = 30 + 10 + 20 = 60 s

3

(a) Average speed = distance/time = 400/60 = 6.67 m s⁻¹

4

(b) Average velocity = displacement/time = 400/60 = 6.67 m s⁻¹

Note: In this case, speed and velocity are equal because the journey was in a straight line in one direction.

Velocity-Time Graphs

Gradient = Acceleration

Positive gradient → speeding up

Negative gradient → slowing down

Zero gradient → constant velocity

Area = Displacement

Use geometry: triangles, rectangles, trapeziums

Area below t-axis → negative displacement

Displacement-Time Graphs

Key Interpretation

Gradient = Velocity

Horizontal line → stationary (v = 0)

Straight line → constant velocity (steeper = faster)

Negative gradient → moving in opposite direction

Curve → changing velocity (draw tangent for instantaneous v)

02Worked Example

From the displacement-time graph below, find the velocity at t = 2 s.

1

Draw a tangent to the curve at t = 2 s (red dashed line touching curve at the red point)

2

Read two points on the tangent: at t = 1 s, s ≈ 0 m; at t = 3 s, s ≈ 40 m

3

Gradient = Δs/Δt = (40 − 0)/(3 − 1) = 40/2 = 20 m s⁻¹

AnswerInstantaneous velocity at t = 2 s is 20 m s⁻¹

03Worked Example

A velocity-time graph shows a car starting from rest, accelerating uniformly for 10 s to 20 m s⁻¹, then travelling at constant velocity for 15 s. Calculate the total displacement.

1

Displacement = area under v-t graph

2

Phase 1 (0-10s): Triangle area = ½ × base × height = ½ × 10 × 20 = 100 m

3

Phase 2 (10-25s): Rectangle area = base × height = 15 × 20 = 300 m

4

Total displacement = 100 + 300 = 400 m

Acceleration-Time Graphs

Key Interpretation

Area under a-t graph = Change in velocity (Δv)

Positive area (above axis) → velocity increases

Negative area (below axis) → velocity decreases

Horizontal line → uniform (constant) acceleration

Graph Relationships Summary

s-t gradient → v | v-t gradient → a | v-t area → s | a-t area → Δv

04Worked Example

A car starting from rest has the acceleration-time graph shown. Find the velocity after 8 seconds.

1

Change in velocity = area under a-t graph

2

Phase 1 (0-4s): Area = 2 × 4 = 8 m s⁻¹

3

Phase 2 (4-8s): Area = 1 × 4 = 4 m s⁻¹

4

Total Δv = 8 + 4 = 12 m s⁻¹

5

Starting from rest (u = 0): Final velocity = 0 + 12 = 12 m s⁻¹

SUVAT Variables

SymbolQuantityUnit

sDisplacementm

uInitial velocitym s⁻¹

vFinal velocitym s⁻¹

aAccelerationm s⁻²

tTimes

Sign Convention

Choose a positive direction and stick to it. For free fall, if you take downward as positive, then g = +9.8 m s⁻². If upward is positive, then g = -9.8 m s⁻².

SUVAT Equations

These equations apply only for uniform (constant) acceleration:

EquationVariables UsedMissing Variable

v = u + a t

v, u, a, ts

s = \frac{( u + v )}{2} t

s, u, v, ta

s = u t + \frac{1}{2} a t^{2}

s, u, a, tv

v^{2} = u^{2} + 2 a s

v, u, a, st

Strategy

List what you know (3 variables) and what you want. Choose the equation that contains your 3 knowns and the unknown.

Worked Example05

A car accelerates uniformly from 8 m s⁻¹ to 20 m s⁻¹ over a distance of 140 m. Calculate the acceleration.

1

List knowns: u = 8 m s⁻¹, v = 20 m s⁻¹, s = 140 m

2

Want: a. Missing: t

3

Use: v² = u² + 2as (equation without t)

4

Rearrange: a = (v² - u²) / 2s

5

a = (20² - 8²) / (2 × 140) = (400 - 64) / 280 = 336 / 280

Answer: a = 1.2 m s⁻²

Worked Example06

A train decelerates uniformly from 30 m s⁻¹ to rest over 45 s. Find: (a) the deceleration, (b) the distance travelled while braking.

1

Knowns: u = 30 m s⁻¹, v = 0, t = 45 s

2

(a) Use v = u + at → 0 = 30 + a(45)

3

a = −30/45 = −0.67 m s⁻² (negative = deceleration)

4

(b) Use s = ½(u + v)t = ½(30 + 0)(45) = 675 m

Free Fall and g

Acceleration Due to Gravity

g = 9.81 m s^{- 2}

(Often approximated as 9.8 or 10 m s⁻²)

Free Fall

An object in free fall is acted on only by gravity. All objects fall with the same acceleration (Galileo).

Gravitational Field Strength

g is also the gravitational field strength.

Units: N kg⁻¹ (equivalent to m s⁻²)

Common scenarios:

Dropped from rest: u = 0

Thrown upward: a = −g

At max height: v = 0

Worked Example07

A stone is dropped from rest from a cliff 45 m high. Taking g = 9.8 m s⁻², calculate: (a) the time to reach the ground, (b) the velocity just before hitting the ground.

1

Take downward as positive. u = 0, s = 45 m, a = g = 9.8 m s⁻²

2

(a) Use s = ut + ½at² → 45 = 0 + ½(9.8)t²

3

45 = 4.9t² → t² = 9.18 → t = 3.03 s

4

(b) Use v = u + at → v = 0 + 9.8 × 3.03 = 29.7 m s⁻¹

Alternative for (b): v² = u² + 2as = 0 + 2(9.8)(45) = 882, so v = 29.7 m s⁻¹

Worked Example08

A ball is thrown vertically upward with velocity 15 m s⁻¹. Taking g = 9.8 m s⁻², find: (a) the maximum height reached, (b) the total time in the air.

1

Take upward as positive: u = +15 m s⁻¹, a = −9.8 m s⁻²

2

At max height, v = 0. Use v² = u² + 2as

3

0 = 15² + 2(−9.8)s → 0 = 225 − 19.6s → s = 11.5 m

4

Time to top: v = u + at → 0 = 15 − 9.8t → t = 1.53 s

5

By symmetry, time down = time up. Total time = 2 × 1.53 = 3.06 s

Bouncing Ball: All Three Graphs

Parabolas - each bounce smaller

Sawtooth - same gradient throughout

Constant a = −g (always)

Key Features

s-t: Inverted parabolas - ball reaches ground, bounces to lower height each time (energy loss)

v-t: Velocity goes negative (falling), instantly reverses positive (bouncing), gradient always = g

a-t: Constant horizontal line at a = −g — gravity acts the same throughout flight

Real-World Applications

Road Safety

Stopping distances depend on thinking distance (proportional to v) and braking distance (proportional to v²).

If speed doubles, braking distance quadruples!

Sports Analysis

Motion analysis of athletes, ball trajectories, and race timing all use these equations.

Engineering

Lift and elevator design, vehicle acceleration performance, runway lengths for aircraft.

Required Practical 3

Determination of g by a free-fall method — typically using light gates or electromagnetic release.

Practice

Question 1: A car accelerates from 12 m s⁻¹ to 28 m s⁻¹ in 8 s. Calculate: (a) the acceleration, (b) the distance travelled during this time.

a)

a = (v - u)/t = (28 - 12)/8 = 16/8 = 2 m s⁻²

b)

s = ½(u + v)t = ½(12 + 28)(8) = ½(40)(8) = 160 m

Practice

Question 2: An object is dropped from a height of 80 m. Taking g = 10 m s⁻², calculate: (a) the time to fall, (b) the speed just before impact.

a)

s = ut + ½at² → 80 = 0 + ½(10)t² → t² = 16 → t = 4 s

b)

v = u + at = 0 + 10(4) = 40 m s⁻¹

Or: v² = u² + 2as = 0 + 2(10)(80) = 1600 → v = 40 m s⁻¹

Practice

Question 3: A motorcycle travelling at 25 m s⁻¹ brakes with a deceleration of 5 m s⁻². Calculate: (a) the time to stop, (b) the braking distance.

a)

v = u + at → 0 = 25 + (-5)t → t = 25/5 = 5 s

b)

v² = u² + 2as → 0 = 625 + 2(-5)s → 10s = 625 → s = 62.5 m

Or: s = ½(u + v)t = ½(25 + 0)(5) = 62.5 m

Summary

Key Concepts

ConceptKey Point

Displacement vs DistanceDisplacement is vector (direction matters), distance is scalar

s-t graph gradientGives velocity

v-t graph gradientGives acceleration

v-t graph areaGives displacement

SUVATOnly for uniform acceleration — list knowns, pick equation

Free falla = g = 9.81 m s⁻² (downward)

SUVAT Equations

v = u + ats = ½(u + v)ts = ut + ½at²v² = u² + 2as