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Travel Graphs & Kinematics GCSE Level

Grade 10 · Algebra & Graphs · Cambridge IGCSE · Age 14–15

Welcome to Travel Graphs & Kinematics!

Travel graphs allow us to describe and analyse motion visually. The distance-time graph shows how far an object has travelled over time, while the velocity-time graph shows how speed changes. Together, they unlock kinematics — the mathematics of motion.

Distance-Time: gradient = speed  |  Velocity-Time: gradient = acceleration, area = distance

Learning Objectives

  • Read and interpret distance-time graphs, finding speed and rest periods
  • Calculate gradient of velocity-time graphs to find acceleration
  • Calculate area under velocity-time graphs to find distance travelled
  • Draw tangents to curved graphs to estimate instantaneous rates
  • Apply kinematics equations: v = u + at, s = ut + ½at², v² = u² + 2as
  • Use the trapezium rule to estimate area under curves
  • Connect distance-time and velocity-time graphs by reasoning about gradients

Distance-Time Graphs

Gradient = speed, horizontal = rest, steeper = faster

Velocity-Time Graphs

Gradient = acceleration, area = distance

Curved Graphs & Kinematics

Tangents, trapezium rule, suvat equations

Worked Examples

6 fully worked mark-scheme examples

Practice 25q

Mixed difficulty questions

Challenge 12q

Multi-step IGCSE problems

Learn 1 — Distance-Time Graphs

A distance-time graph plots distance (y-axis, usually metres or km) against time (x-axis, usually seconds, minutes or hours). The shape of each segment tells a story about the object's motion.

Gradient = Speed

Speed = rise ÷ run = change in distance ÷ change in time = Δd / Δt
To find speed from a distance-time graph: pick two clear points on the line, read off the coordinates, then calculate (y₂ − y₁) ÷ (x₂ − x₁). The units of the gradient are the units of the y-axis divided by the units of the x-axis — e.g. km/h or m/s.

Interpreting Segments

Segment ShapeMeaning
Upward sloping straight lineMoving away from start at constant speed
Horizontal (flat) lineStationary — not moving (speed = 0)
Steeper upward slopeFaster speed than a shallower slope
Downward sloping lineReturning towards the starting point
Curved lineNon-constant speed (accelerating or decelerating)

Average Speed

Average speed = total distance travelled ÷ total time taken
Warning: Average speed is NOT the gradient of the straight line joining start to end point on a graph — that gives the average velocity. If the object went away and came back, total distance includes both journeys.

Reading Values from a Graph

Distance from start at a given time: read vertically from the time value up to the graph, then across to the y-axis.
Time of rest: identify horizontal sections; note the time interval.
Total distance: add up all upward and downward movements (not just the final position).

Negative Gradient — Returning

A negative gradient on a distance-time graph means the object is returning towards its starting point (distance from start is decreasing). Speed on the return journey = |gradient| (take the absolute value since speed is always positive).

Learn 2 — Velocity-Time Graphs

A velocity-time (v-t) graph plots velocity (y-axis, m/s) against time (x-axis, seconds). From this graph we can extract both acceleration and distance information.

Gradient = Acceleration

Acceleration = change in velocity ÷ change in time = Δv / Δt  (units: m/s²)
GradientMeaning
PositiveAcceleration (speeding up)
Zero (horizontal)Constant velocity (zero acceleration)
NegativeDeceleration (slowing down)

Area Under Graph = Distance

Distance = area under the velocity-time graph
Rectangle (constant velocity): Area = base × height = time × velocity
Triangle (uniform acceleration from/to zero): Area = ½ × base × height
Trapezium (general): Area = ½(a + b) × h, where a and b are the parallel sides (velocities) and h is the time interval
Total distance = sum of ALL areas. If part of the graph is below the x-axis (negative velocity), treat that area as positive when calculating total distance, but negative when calculating displacement (net change in position).

Typical v-t Graph Shape

A common IGCSE question gives a trapezium shape: the object accelerates from rest (triangle on left), travels at constant velocity (rectangle in middle), then decelerates to rest (triangle on right). Calculate each area separately, then add.

Learn 3 — Curved Graphs and Estimation

Curved Distance-Time Graphs

A curve on a d-t graph means the speed is not constant — the object is accelerating or decelerating. To find the instantaneous speed at a particular point, draw a tangent to the curve at that point and calculate its gradient.
Tangent vs Chord: A tangent JUST touches the curve at one point. A chord joins two points on the curve. Only the tangent gives the instantaneous speed.

Curved Velocity-Time Graphs

A curve on a v-t graph means acceleration is non-constant. To find instantaneous acceleration, draw a tangent and find its gradient. To find distance, estimate the area using the trapezium rule:
Area ≈ ½h[(y₀ + yₙ) + 2(y₁ + y₂ + … + yₙ₋₁)] where h = width of each strip

Kinematics Equations (Constant Acceleration)

These four equations apply only when acceleration is uniform (constant):

EquationVariables connected
v = u + atv, u, a, t
s = ut + ½at²s, u, a, t
v² = u² + 2asv, u, a, s
s = ½(u + v)ts, u, v, t
Variables: s = displacement (m), u = initial velocity (m/s), v = final velocity (m/s), a = acceleration (m/s²), t = time (s)

Connecting the Two Graphs

Given a d-t graph → the gradient at each point gives the v-t graph.
Given a v-t graph → the area up to each time gives the d-t graph.
If d-t graph is a straight line → v-t graph is a horizontal line (constant velocity).
If d-t graph curves upward → v-t graph slopes upward (increasing velocity).

Worked Examples

Example 1 — Reading Speed from a Distance-Time Graph

A car travels from home. At t = 0 it is at distance 0. At t = 2h it is at 80 km. At t = 3h it stops and remains at 80 km until t = 4h. It then returns, reaching 0 km at t = 6h. Find: (a) speed in first 2 hours, (b) speed on return journey.

Part (a): Gradient of first segment = Δd / Δt = (80 − 0) / (2 − 0) = 40 km/h
Part (b): Return journey: from (4h, 80km) to (6h, 0km). Gradient = (0 − 80) / (6 − 4) = −40 km/h. Speed = |−40| = 40 km/h
Answers: (a) 40 km/h (b) 40 km/h. Note: total distance = 80 + 80 = 160 km in 6h → average speed = 160/6 ≈ 26.7 km/h ≠ gradient of the straight line from start to end.

Example 2 — Total Distance from a Velocity-Time Graph

A cyclist: accelerates from 0 m/s to 12 m/s in the first 4 s, maintains 12 m/s for 6 s, then decelerates to 0 m/s in 3 s. Calculate the total distance travelled.

Area 1 (triangle, 0–4s): ½ × 4 × 12 = 24 m
Area 2 (rectangle, 4–10s): 6 × 12 = 72 m
Area 3 (triangle, 10–13s): ½ × 3 × 12 = 18 m
Total distance = 24 + 72 + 18 = 114 m
Total distance = 114 m

Example 3 — Finding Acceleration

A train increases its speed from 10 m/s to 34 m/s over 8 seconds. Calculate the acceleration.

Acceleration = (v − u) / t = (34 − 10) / 8 = 24 / 8 = 3 m/s²
Acceleration = 3 m/s²

Example 4 — Sketch v-t from Description

A ball is thrown upward at 20 m/s, decelerates at 10 m/s² until it stops momentarily, then accelerates downward at 10 m/s². Sketch the v-t graph for 4 seconds.

At t = 0: v = 20 m/s (upward, positive). Decelerates at 10 m/s².
Reaches v = 0 at t = 20/10 = 2 s. Then crosses x-axis (velocity becomes negative = downward).
At t = 4 s: v = 0 − 10×2 = −20 m/s. The v-t graph is a straight line from (0, 20) to (4, −20), passing through (2, 0).
Straight line with gradient −10, from (0, 20) crossing zero at t = 2 s, reaching −20 m/s at t = 4 s.

Example 5 — Trapezium Rule for Area

A curved v-t graph has velocities recorded at t = 0, 2, 4, 6, 8 s as: 0, 6, 10, 12, 13 m/s. Use the trapezium rule to estimate distance.

h = 2 s (equal intervals). Apply: Area ≈ ½ × h × [(y₀ + y₄) + 2(y₁ + y₂ + y₃)]
= ½ × 2 × [(0 + 13) + 2(6 + 10 + 12)]
= 1 × [13 + 2(28)] = 1 × [13 + 56] = 69 m
Estimated distance ≈ 69 m

Example 6 — Kinematics Equation

A car starts from rest and accelerates at 2.5 m/s². Find: (a) its speed after 8 s, (b) the distance after 8 s, (c) the speed when it has travelled 180 m.

(a) v = u + at = 0 + 2.5 × 8 = 20 m/s
(b) s = ut + ½at² = 0 + ½ × 2.5 × 64 = 80 m
(c) v² = u² + 2as = 0 + 2 × 2.5 × 180 = 900, so v = √900 = 30 m/s
(a) 20 m/s   (b) 80 m   (c) 30 m/s

Common Mistakes

Mistake 1 — Confusing the two graphs: On a distance-time graph, gradient = speed. On a velocity-time graph, gradient = acceleration. Students often read a v-t graph's gradient as speed, or confuse area on a d-t graph as meaning something. Area has meaning only on a v-t graph.
Mistake 2 — Using total distance / total time for instantaneous speed: That formula gives the average speed for the whole journey. For speed at a specific instant or for a specific segment, you must use the gradient of that specific segment.
Mistake 3 — Negative area in v-t graphs: If the graph dips below the x-axis, velocity is negative (object moving backwards). For total distance, treat all areas as positive. For displacement (net change in position), subtract the area below the axis. Confusing these two concepts is a very common error.
Mistake 4 — Drawing a chord instead of a tangent: To find instantaneous speed/acceleration from a curved graph, you must draw the tangent — a line that just touches the curve at one point without crossing it. A chord joins two points on the curve and gives the average rate over that interval, not the instantaneous rate.
Mistake 5 — Forgetting units of gradient: Gradient units = (y-axis units) ÷ (x-axis units). If velocity is in m/s and time in seconds, gradient = m/s ÷ s = m/s². If speed is in km/h and time in hours, gradient = km/h ÷ h = km/h². Always state the units in your answer.

Key Formulas

FormulaWhat it givesStatus
Speed = Δd / ΔtSpeed from d-t gradientGiven in exam
Acceleration = Δv / ΔtAcceleration from v-t gradientMust memorise
Distance = area under v-t graphDistance from v-t graphMust memorise
Area of trapezium = ½(a+b)hArea of trapezoidal segmentGiven in exam
v = u + atFinal velocityGiven in exam
s = ut + ½at²DisplacementGiven in exam
v² = u² + 2asVelocity without timeGiven in exam
s = ½(u+v)tDisplacement (avg velocity × time)Must memorise
Trapezium rule: ½h[(y₀+yₙ)+2(y₁+…+yₙ₋₁)]Estimated area under curveGiven in exam
Average speed = total distance / total timeOverall average speedMust memorise
Exam tip: The kinematics equations (suvat) are provided on the formula sheet in Cambridge IGCSE examinations. However, you must know when to apply them — only when acceleration is constant. For variable acceleration, use the graph methods.

Velocity-Time Graph Visualiser

Adjust the sliders to set the velocity at three points. The canvas draws the v-t graph, shades the area (distance), and calculates the gradient (acceleration) for each segment.

Exercise 1 — Reading Distance-Time Graphs

A person walks from home. At t=0, d=0. At t=1h, d=5km. Stays until t=2h. Returns, reaching d=0 at t=4h.

Q1. What is the speed during the first hour? (km/h)

Q2. How long did the person rest? (hours)

Q3. What is the speed on the return journey? (km/h)

Q4. What is the total distance travelled? (km)

Q5. What is the average speed for the whole journey? (km/h)

Q6. At what distance is the person at t = 3h? (km)

Q7. What does the flat section of the d-t graph tell us?

Q8. If the person had walked back at 3 km/h, how long would the return journey take? (hours)

Exercise 2 — Velocity-Time Graph Calculations

A bus: accelerates from 0 to 18 m/s in 6 s, then travels at 18 m/s for 10 s, then decelerates to 0 in 4 s.

Q1. What is the acceleration in the first 6 s? (m/s²)

Q2. What is the deceleration in the last 4 s? (m/s²) (give as positive)

Q3. Distance covered during acceleration phase? (m)

Q4. Distance covered during constant speed phase? (m)

Q5. Distance covered during deceleration phase? (m)

Q6. Total distance covered? (m)

Q7. Total time of journey? (s)

Q8. Average speed for whole journey? (m/s, 2dp)

Exercise 3 — Kinematics Equations

Q1. A car starts from rest and accelerates at 4 m/s². Find its speed after 5 s. (m/s)

Q2. A train travels at 30 m/s and brakes with deceleration 2 m/s². How long to stop? (s)

Q3. Using v² = u² + 2as: u = 0, a = 3 m/s², s = 48 m. Find v. (m/s)

Q4. Object starts at 10 m/s, accelerates at 2 m/s² for 6 s. Find distance. (m)

Q5. A ball is thrown upward at 15 m/s. Taking g = 10 m/s², when does it reach maximum height? (s)

Q6. Using s = ½(u+v)t: u = 0, v = 20 m/s, t = 8 s. Find s. (m)

Q7. A runner starts at 2 m/s and reaches 8 m/s in 3 s. Find acceleration. (m/s²)

Q8. Car decelerates from 25 m/s to rest over 50 m. Find deceleration. (m/s²)

Exercise 4 — Trapezium Rule & Mixed Areas

Q1. v-t graph: trapezium with u=4, v=10, t=5. Find area (distance). (m)

Q2. Trapezium rule, h=2, y values: 0, 8, 14, 18, 20. Estimate area. (m)

Q3. A v-t graph has a triangle (base=4s, height=12 m/s) and a rectangle (4s × 12 m/s). Total distance? (m)

Q4. Trapezium rule, h=3, y values: 0, 5, 8, 9. Estimate area. (m)

Q5. A car travels at 20 m/s for 8 s, then decelerates to 0 in 5 s. Total distance? (m)

Q6. v-t graph: velocities at t=0,1,2,3,4 are 2,5,7,8,8 m/s. Use trapezium rule (h=1). Estimate distance. (m)

Q7. Object moves at −3 m/s for 4 s (below x-axis). What is the distance travelled? (m)

Q8. Two sections: triangle (base=3, height=6) below x-axis; rectangle (4×5) above. Total distance? (m)

Exercise 5 — Sketching and Connecting Graphs

Q1. A d-t graph is horizontal for 3 s. What is the speed during this time? (m/s)

Q2. A d-t graph has a straight line with gradient 8 m/s. What does the corresponding v-t graph look like?

Q3. The v-t graph shows constant acceleration from 0 to 15 m/s in 5 s. What is the shape of the d-t graph?

Q4. Speed = 12 m/s for 7 s, then 0 m/s. What is the gradient of the d-t graph in the moving phase? (m/s)

Q5. A v-t graph shows deceleration from 20 m/s to 0 in 10 s. What is the acceleration? (m/s², negative)

Q6. What does a downward slope on a d-t graph mean about the object's direction?

Q7. An object has a curved d-t graph with increasing gradient. What does the v-t graph look like?

Q8. A v-t graph is horizontal at 5 m/s. What is the shape of the corresponding d-t graph?

Practice — 25 Questions

🔵 P1. Gradient of d-t graph = (8−0)/(2−0). Find speed. (m/s)

🔵 P2. A flat section on a d-t graph lasts 5 s. Distance covered? (m)

🔵 P3. v = u + at: u=0, a=5, t=4. Find v. (m/s)

🔵 P4. Area of triangle on v-t: base=6s, height=12 m/s. Distance? (m)

🔵 P5. Acceleration = (20−5)/3. Calculate. (m/s²)

🔵 P6. s = ut + ½at²: u=0, a=4, t=3. Find s. (m)

🔵 P7. Trapezium area: ½(6+14)×5. Find value. (m)

🟢 P8. A car travels at 15 m/s for 8 s, then at 25 m/s for 4 s. Total distance? (m)

🟢 P9. v²=u²+2as: u=0, a=6, s=75. Find v. (m/s)

🟢 P10. Object decelerates from 18 m/s to 0 in 9 s. Find deceleration (positive). (m/s²)

🟢 P11. Trapezium rule h=2: y values 0, 4, 7, 9, 10. Estimate area. (m)

🟢 P12. Average speed = total distance / total time. Journey: 60 km in 1.5 h. Average speed? (km/h)

🟢 P13. s = ½(u+v)t: u=5, v=25, t=10. Find s. (m)

🟢 P14. v-t graph: accelerates 0→10 m/s in 3s, constant 10 m/s for 5s, decelerates to 0 in 2s. Total distance? (m)

🟢 P15. A train starts at 20 m/s and accelerates at 1.5 m/s² for 10 s. Final speed? (m/s)

🟢 P16. Using v²=u²+2as to find s: u=10, v=0, a=−5. Find s. (m)

🟢 P17. A bus journey: 0→15 m/s in 5 s (triangle). Find area under v-t for this section. (m)

🟢 P18. D-t graph: from (0,0) to (4,20). Gradient = speed. (m/s)

🟢 P19. Object below x-axis on v-t graph for 3 s at −4 m/s. Distance travelled? (m)

🟢 P20. Acceleration = gradient of v-t. From (2,5) to (8,29). Find a. (m/s²)

🟢 P21. s = ut + ½at²: u=6, a=−2, t=3. Find s. (m)

🟢 P22. Total distance: triangle (b=5,h=10) + trapezium ½(10+16)×4 + triangle (b=3,h=16). Total? (m)

🟢 P23. Object at constant velocity 12 m/s for 8 s. What is area under v-t? (m)

🟢 P24. Using v = u + at: v=40, u=10, a=5. Find t. (s)

🟢 P25. Trapezium rule h=1, y: 0,3,5,6,6.5,7. Estimate area (units²).

Challenge — 12 Multi-Step Questions

C1. A cyclist accelerates from 0 to 8 m/s in 4 s, travels at 8 m/s for 10 s, then decelerates to 3 m/s in 5 s. Find total distance. (m)

C2. A car decelerates from 30 m/s. If it travels 90 m while stopping, find the deceleration. (m/s²)

C3. Trapezium rule: h=3, y₀=0, y₁=9, y₂=15, y₃=18, y₄=19. Estimate total area. (units)

C4. Two objects start together. Object A: constant 10 m/s. Object B: accelerates at 2 m/s² from rest. When does B overtake A? (s)

C5. A stone is dropped from 80 m (a = 10 m/s², u = 0). How long to hit ground? (s)

C6. A v-t graph: from t=0 to t=6 it rises from 5 to 17 m/s, then falls from 17 to 5 m/s in the next 6 s. Find total distance (trapezium). (m)

C7. Object at 20 m/s decelerates at 4 m/s². Find distance when v = 8 m/s. (m)

C8. A train: accelerates from 0 to v m/s in 30 s (distance 450 m). Find v. (m/s)

C9. Find displacement when above x-axis area = 50 m and below x-axis area = 14 m. Displacement? (m)

C10. Curved v-t: tangent at t=4s has gradient 3.5 m/s². What is the instantaneous acceleration? (m/s²)

C11. A ball thrown upward at 25 m/s (g=10 m/s²). Find max height. (m)

C12. v-t graph, trapezium rule h=2: velocities at t=0,2,4,6,8,10 are 4,7,9,10,10,9 m/s. Estimate area. (m)

Exam Style — 5 Structured Questions

Q1. A lorry starts from rest and accelerates uniformly to 18 m/s in 12 seconds. It then maintains this speed for 20 seconds before braking uniformly to rest in 8 seconds.
(a) Calculate the acceleration in the first phase. (m/s²)
(b) Calculate the total distance. (m)
(c) Find the average speed for the whole journey. (m/s, 2dp)

Q2. A particle moves so that its velocity at time t is given by the v-t graph with points: (0,0), (4,16), (10,16), (14,0).
(a) Find the acceleration in the first 4 s. (m/s²)
(b) Find the total distance. (m)
(c) Find the deceleration in the last section. (m/s²)

Q3. A stone is thrown upward with initial velocity 30 m/s. Take g = 10 m/s² (deceleration).
(a) Find the time to reach maximum height. (s)
(b) Find the maximum height. (m)
(c) Find the speed when it returns to start. (m/s)

Q4. The table shows velocity at t = 0, 3, 6, 9, 12 s: 2, 8, 12, 14, 15 m/s. Use the trapezium rule (h = 3).
(a) Write out the trapezium rule formula for this data. (show numbers)
(b) Calculate the estimated distance. (m)

Q5. Two cars start from the same point. Car A travels at a constant 20 m/s. Car B starts from rest and accelerates at 5 m/s².
(a) When does B reach the same speed as A? (s)
(b) How far ahead is A when B reaches A's speed? (m)
(c) When does B catch A? (s)