PhysicsExam code: 4PH1

Movement & Position

Forces & Motion

PractiseFlashcards

What the area means

  • The area enclosed between a velocity-time line and the time axis equals the distance that has been covered during the interval (or, equivalently, the magnitude of the displacement if the motion runs in a single direction)
  • This is true because area = velocity × time on every thin vertical strip under the line, and velocity × time = distance moved in that strip

Splitting a multi-stage motion into shapes

  • A constant-velocity section is a horizontal line, so the area under it is a rectangle:

rectangle area = base × height

  • A section of constant acceleration or deceleration is a straight slope, so the area under it (down to the time axis) is a triangle:

triangle area = ½ × base × height

  • A section that combines a non-zero starting velocity with an acceleration gives a trapezium; split it into a rectangle and a triangle and add the two
  • The total distance for a multi-stage motion is found by adding together the area of every enclosed region one stage at a time
Velocity-time graph with the region under the line divided into a triangle beneath the sloping part labelled ½ × base × height and a rectangle beneath the flat part labelled base × height, with a note that distance = area
Source: Area under velocity graph by Save My Exams

Common exam question

Distance from the area under a velocity–time graph

Question: Calculate the distance travelled, or the braking distance, over a stated interval of a velocity–time graph, or show that it has a given value (3–5 marks).

Asked in 6 of the 24 papers. The first mark is for the idea that distance = area under the line, stated or implied by your working. The method marks are then the shapes: a triangle (½ × base × height) under a sloping section and a rectangle under a flat one, added together; for a curve, count the squares or split the area into trapeziums, and the answer is accepted within a range.

Take the area only over the interval asked for. A braking distance is the triangle under the deceleration alone: adding the rectangle for the reaction time gives the whole stopping distance and lost a mark. Using v² = u² + 2as with the acceleration from the earlier part is accepted as an alternative, with follow-through on that value.

Worked example

Finding distance from the area under a velocity–time graph

A velocity–time graph shows an object accelerating uniformly from rest (0 m/s) to 6.0 m/s over 4.0 s, then the recording stops.

A velocity-time graph on a numbered grid, time in seconds from 0 to 5 along the bottom and velocity in metres per second from 0 to 8 up the side. A straight line rises from the origin to a marked point at 4 seconds and 6 metres per second, and the triangle between that line, the time axis and the vertical at 4 seconds is shaded.

Solution:

  • The region under the line is a triangle (straight slope from zero)
  • Area of triangle = ½ × base × height = ½ × 4.0 × 6.0
  • Distance = 12 m

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