Describing motion along a line velocity–time graphs and uniform acceleration

SPEC 4.5.6.1.5 • TOPIC 5 FORCES — PAPER 2

V-t Graphs & Uniform Acceleration

Two equations, two graph readings. Gradient gives acceleration. Area gives distance (HT). Don't confuse this graph with d-t.

What You'll Learn

The motion topic continues. Velocity-time graphs unlock two readings from a single line — and there's a new equation on the equation sheet for uniform acceleration.

Lesson Goals

Define acceleration, deceleration, uniform acceleration

Apply a = Δv ÷ t (recall) to calculate acceleration

Estimate the magnitude of everyday accelerations

Read a v-t graph: gradient = acceleration; area under = distance HT

Apply v² − u² = 2 a s (equation sheet) for uniform acceleration

Interpret a v-t graph for an object reaching terminal velocity ⚡ Physics only

Key Definitions

Acceleration

Acceleration is the rate of change of velocity. Vector quantity (has direction). Measured in m/s². An acceleration of 3 m/s² means the velocity increases by 3 m/s every second.

Deceleration

An object that is slowing down is decelerating. Mathematically, this is a negative acceleration. In words on a graph: "decelerating at 2 m/s²" = "accelerating at −2 m/s²".

Uniform acceleration

Uniform acceleration = constant acceleration. The velocity changes by the same amount each second. On a v-t graph, uniform acceleration shows as a straight line (not a curve).

Equation 1 — a = Δv ÷ t

Recall equation. Not on the equation sheet. Use this for average acceleration over a time interval.

Acceleration = change in velocity ÷ timea = Δv / ta — m/s²Δv — m/st — s

What "Δv" means

Δv = final velocity − initial velocity = v − u. If the object speeds up, Δv is positive → positive a. If it slows down, Δv is negative → negative a (deceleration). Sign tells you direction of the change.

Rearranged forms

Δv = a × t | t = Δv ÷ a

Estimating Everyday Accelerations

The spec specifically requires you to estimate the magnitude of everyday accelerations. These typical values are your reference points.

~1

m/s² — train braking

~3

m/s² — car from traffic lights

~5

m/s² — sprinter at start

9.8

m/s² — free fall near Earth

~10

m/s² — sports car (0–60 mph in ~3 s)

~30

m/s² — roller-coaster peak (~3g)

The key value to remember

Free-fall acceleration ≈ 9.8 m/s² near the Earth's surface (this is g — the same value as the gravitational field strength). For estimation, sometimes rounded to 10 m/s². The question will tell you which to use.

ESUA Worked Example

A car travelling at 20 m/s brakes to a slower speed of 5 m/s over a time of 3.0 s.

Calculate the car's average acceleration. State whether it is decelerating. 3 MARKS

E

Equation: a = Δv ÷ t = (v − u) ÷ t

S

Substitute: Δv = 5 − 20 = −15 m/s; t = 3.0 s; a = −15 ÷ 3.0

U

Unit: metres per second squared (m/s²)

A

Answer: a = −5 m/s² (the car decelerates at 5 m/s²). The negative sign confirms the car is slowing down.

The Velocity-Time Graph

A v-t graph shows how an object's velocity changes over time. Convention: time on x-axis, velocity on y-axis. Two readings — pulled from the same line — give you two different physical quantities.

The two readings

Gradient (slope) of the line = acceleration. A steeper line = greater acceleration. A horizontal line = zero acceleration (constant velocity). A line sloping downward = deceleration.

Area under the line = distance travelledHT. To find the area, identify the geometric shape (rectangle, triangle, trapezium) and calculate it.

Plotting a v-t graph from data (spec skill)

The spec also requires you to draw a v-t graph from measurements. Method:

Put time on the x-axis, velocity on the y-axis — with a scale and a unit on each

Plot each (time, velocity) measurement as a small cross or dot

Draw a straight line or smooth curve of best fit through the points — not a zigzag connecting them

Don't confuse v-t with d-t

On a d-t graph: gradient = SPEED. On a v-t graph: gradient = ACCELERATION, and area = distance. Same shape lines mean completely different things. Always check which graph you're reading.

The Four Basic V-t Shapes

Recognise these on sight. Each shape tells a story about the object's motion.

Constant velocity
a = 0 — Constant velocity a = 0

Uniform acceleration
from rest — Uniform acceleration from rest

Uniform deceleration
to rest — Uniform deceleration to rest

Accelerate then constant two phases

Gradient interpretation

Steeper positive slope = greater acceleration. Steeper negative slope = greater deceleration. Horizontal segment = constant velocity (zero acceleration). The line never has to start at the origin — an object can already be moving at t = 0.

Area Under V-t = Distance HT

For Higher Tier, the area enclosed under a v-t graph equals the distance travelled (or displacement, if direction matters). To find the area, identify the geometric shape under each segment of the line and use the right formula.

RECTANGLE
constant velocity — RECTANGLE constant velocity

TRAPEZIUM u to v

Method

Identify the shape of each segment (rectangle, triangle, trapezium, or count squares for curves)

Read the values for base (time) and height (velocity) from the graph axes — use the actual scales

Apply the correct formula and sum the areas of all segments

Counting squares — for curves & irregular shapes HT

For non-uniform acceleration the v-t line is curved, so the area underneath isn't a standard rectangle or triangle. The spec method is to count squares:

Count every whole square that sits entirely under the curve

For partial squares: count as a full square if half or more is under the curve; ignore if less than half is under

Find the value of one square from the axes: 1 square = (x-step) × (y-step). E.g., if each square is 1 s × 2 m/s, then 1 square represents 2 m

Multiply: total squares × value of one square = distance travelled

ESUA Worked Example — From a V-t Graph

(a)

3 MARKS

(b)

(a)

Method: a = gradient = Δv ÷ t

Substitute: Δv = 12 − 0 = 12 m/s; t = 4 − 0 = 4 s; a = 12 ÷ 4 (M1, M1)

Answer: a = 3 m/s² (A1)

(b)

Method: distance = area under v-t = area of triangle = ½ × base × height (M1)

Substitute: area = ½ × 4 × 12 (M1)

Answer: distance = 24 m (A1)

Equation 2 — v² − u² = 2 a s Equation Sheet

The second motion equation. On the equation sheet — but you must know how to apply it. Applies only to uniform (constant) acceleration.

Final velocity² − Initial velocity² = 2 × acceleration × distancev² − u² = 2 a sv — final, m/su — initial, m/sa — m/s²s — m

When to use it

Use v² − u² = 2as when the problem gives you (or asks for) a relationship between distance, velocities, and acceleration — but NOT time. If time is given or asked for, use a = Δv ÷ t instead.

Uniform acceleration only

This equation is valid only when the acceleration is constant. If the v-t graph is a curve (not a straight line), the acceleration is changing and this equation doesn't apply. Use area under the v-t graph (HT) instead.

ESUA Worked Example — v² − u² = 2 a s

A car accelerates uniformly from rest. It reaches a velocity of 25 m/s after travelling 200 m.

Calculate the car's acceleration. 4 MARKS

E

Equation (sheet): v² − u² = 2 a s, rearranged for a: a = (v² − u²) ÷ (2s)

S

Substitute: u = 0 (from rest); v = 25 m/s; s = 200 m

a = (25² − 0²) ÷ (2 × 200) = 625 ÷ 400

U

Unit: metres per second squared (m/s²)

A

Answer: a ≈ 1.6 m/s² (2 s.f.)

Free Fall & Terminal Velocity ⚡ Physics only

An object falling near the Earth's surface accelerates at g ≈ 9.8 m/s² due to gravity (in vacuum, with no air resistance). When falling through a fluid like air, this changes.

Falling through a fluid — three stages

Initial: only gravity acts (or air resistance is small). Object accelerates at ≈ g.

Speed increases → air resistance increases. The resultant force (weight − drag) decreases, so the acceleration decreases.

Air resistance = weight. Resultant force = 0. Acceleration = 0. Velocity is now constant — this is terminal velocity.

Force-balance explanation

Terminal velocity is reached when upward air resistance equals the downward weight. With zero resultant force, there's zero acceleration (Newton's 2nd Law) — so the object continues at constant velocity. A skydiver in freefall reaches a terminal velocity of about 55 m/s; with parachute open, it drops to about 7 m/s (larger surface area → more air resistance).

Exam-Style Multi-Part Question

(a)

1 MARK

(b)

3 MARKS

(c)

3 MARKS · HT

(a)

The cyclist moves at constant velocity (the line is horizontal at 8 m/s, so acceleration = 0). (1)

(b)

Method: a = Δv ÷ t (gradient) (M1)

Substitute: Δv = 8 − 0 = 8 m/s; t = 5 s; a = 8 ÷ 5 (M1)

Answer: a = 1.6 m/s² (A1)

(c)

Method: total distance = area under v-t = sum of three shapes (M1)

Triangle 1 (0–5 s): ½ × 5 × 8 = 20 m

Rectangle (5–15 s): 10 × 8 = 80 m

Triangle 2 (15–19 s): ½ × 4 × 8 = 16 m (M1)

Answer: total = 20 + 80 + 16 = 116 m (A1)

Examiner Alerts & Weak vs Strong

The #1 confusion — v-t vs d-t

On a v-t graph: gradient = acceleration, area = distance. On a d-t graph: gradient = speed; the area underneath has no physical meaning. Same-shaped lines on the two graphs mean completely different things. Always check the axes labels first.

Question: "Calculate the distance travelled in the first 5 s. (HT)" — referring to a v-t graph.

✗ Weak

"distance = gradient × time = 1.6 × 5 = 8 m"Used the GRADIENT (which is the acceleration) for distance. Wrong operation. 0/3.

✓ Strong

"distance = area under graph = ½ × 5 × 8 = 20 m"Method named (area), shape identified (triangle), substitution, answer with unit. 3/3.

When NOT to use v² − u² = 2as

Only use this equation for uniform (constant) acceleration — straight-line v-t graphs. If the v-t graph is curved, the acceleration is changing — use area under the curve instead (count squares for irregular shapes).

Common Errors

Students often writeCorrect versionWhy it matters

Area under d-t graph = distanceArea under v-t graph = distance. Area under d-t graph has no physical meaningWrong graph type completely

Use v² − u² = 2as for non-uniform accelerationOnly valid for uniform (constant) acceleration — straight line on v-t graphEquation misapplication = wrong answer

"Acceleration" stated without sign or directionFor deceleration, state either "−5 m/s²" or "5 m/s² (decelerating)"Mark schemes accept the sign or the word

Used initial velocity in s = vt for accelerating motionFor uniform a: use average velocity (u+v)/2 in s = (avg v) × t, or use the area-under-graph methodInitial v alone undercounts distance

Reading the scale wrong (e.g., taking each square as 1 unit when scale = 2)Always read the actual numbered axis values; multiply by the scale factorOff by factor of 2, 5, or 10

Answer "1.6" with no unit"1.6 m/s²""Calculate" answers always need a unit

Checkpoint

You can now

Define acceleration, deceleration, and uniform acceleration

Apply a = Δv ÷ t to calculate average acceleration

Estimate typical everyday accelerations (free-fall 9.8 m/s², car ~3 m/s², sprinter ~5 m/s²)

Read a v-t graph: gradient = acceleration; area under = distance (HT)

Calculate areas of rectangles, triangles, and trapeziums under a v-t graph

Apply v² − u² = 2as for uniform acceleration

Interpret a terminal-velocity v-t graph in terms of force balance (physics only)

The one habit

Every v-t question starts with the same check: am I asked for acceleration or distance? Acceleration → gradient. Distance → area. Reach for the right operation and the answer follows.

Next lesson

5.10 — Forces, accelerations and Newton's Laws of motion & RP7. Newton's three laws, F = m a (recall), inertia, inertial mass (HT), and the required practical investigating acceleration vs force/mass.