Velocity–Time Graph

\( \text{Acceleration}=\text{gradient},\quad \text{Distance travelled}=\text{area under the graph} \)
Measures GCSE

A car travels at 18 m/s for 10 s. Find distance travelled.

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( Area under graph = rectangle base×height. \)

Explanation

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Statement

A velocity–time graph shows how velocity changes with time. Two key facts are used:

\[ \text{Acceleration} = \text{gradient of the velocity–time graph} \]

\[ \text{Distance travelled} = \text{area under the velocity–time graph} \]

The gradient gives acceleration because it measures the rate of change of velocity. The area under the graph gives distance because velocity multiplied by time equals distance.

Why it’s true

  • Acceleration is defined as change in velocity ÷ time. On the graph, this is rise ÷ run, i.e. the gradient.
  • Distance = speed × time. On the graph, velocity is the vertical axis and time is the horizontal axis, so the area beneath the curve represents this product.
  • For straight-line sections, this is a triangle or rectangle; for curved sections, it is found by approximation or integration (beyond GCSE).

Recipe (how to use it)

  1. Identify whether you are asked for acceleration (look for gradient) or distance (look for area).
  2. For acceleration, find the slope of the line: \(\Delta v / \Delta t\).
  3. For distance, compute the area under the line segment (triangle, rectangle, trapezium, or combination).

Spotting it

Questions with velocity–time graphs often say: “find the acceleration” or “calculate the distance travelled after … seconds”.

Common pairings

  • Acceleration linked with equations of motion (SUVAT).
  • Distance travelled compared with displacement (careful distinction at higher levels).

Mini examples

  1. Given: Straight line from (0,0) to (4,8). Acceleration: (8–0)/(4–0)=2 m/s².
  2. Given: Constant velocity line at v=6 from t=0 to t=5. Distance: Area = 6×5=30 m.

Pitfalls

  • Mixing up gradient and area — check what the question is asking.
  • Forgetting units (e.g. m/s² for acceleration, metres for distance).
  • Calculating only one shape’s area when several need to be combined.

Exam strategy

  • Label axes and note units before starting.
  • Divide the shape under the graph into rectangles and triangles.
  • Always write down formula used: gradient = rise/run, area = base×height or ½×base×height.

Summary

On a velocity–time graph, gradients give accelerations and areas give distances. This is one of the most important applied uses of graphs in GCSE science and maths, connecting algebra, geometry, and kinematics.

Worked examples

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  1. A particle increases velocity from 0 m/s to 10 m/s in 5 s. Find the acceleration.
    1. \( Gradient = (10-0)/(5-0) = 10/5 \)
    2. \( Acceleration = 2 m/s² \)
    Answer: 2 m/s²
  2. A car travels at constant 6 m/s for 8 s. Find the distance travelled.
    1. \( Area = velocity × time = 6 × 8 \)
    2. \( Distance = 48 m \)
    Answer: 48 m
  3. A velocity–time graph shows a line from (0,0) to (4,12). Find the acceleration.
    1. \( Gradient = (12-0)/(4-0)=12/4 \)
    2. \( Acceleration = 3 m/s² \)
    Answer: 3 m/s²
  4. A bike moves at constant velocity of 5 m/s for 12 s. Find distance travelled.
    1. \( Distance = 5 × 12 = 60 m \)
    Answer: 60 m
  5. A velocity–time graph shows line (0,0) to (6,18). Find acceleration.
    1. \( Gradient = 18/6=3 \)
    2. \( Acceleration = 3 m/s² \)
    Answer: 3 m/s²
  6. A car moves at 20 m/s for 7 s. Find distance travelled.
    1. \( Area = 20 × 7 = 140 m \)
    Answer: 140 m
  7. Velocity increases uniformly from 0 to 15 m/s in 10 s. Find acceleration.
    1. \( Gradient = 15/10=1.5 \)
    2. \( Acceleration = 1.5 m/s² \)
    Answer: 1.5 m/s²
  8. A graph shows velocity constant at 8 m/s for 15 s. Find distance.
    1. \( Area=8×15=120 m \)
    Answer: 120 m
  9. A velocity–time graph rises from (0,0) to (5,20). Find acceleration.
    1. \( Gradient = 20/5=4 \)
    2. \( Acceleration=4 m/s² \)
    Answer: 4 m/s²
  10. A bus travels at 12 m/s for 10 s. Find the distance.
    1. \( Area=12×10=120 m \)
    Answer: 120 m