Velocity–time graphs

v–t graph: slope = acceleration; area under the curve = displacement.

Velocity–time graphs0 min · Free lecture

In this lesson

  • v–t graph: slope = acceleration; area under the curve = displacement.

The area under v–t is the whole story of how far you went.

Slope of v–t = acceleration. Area between the curve and the time axis = displacement (signed: below the axis is negative displacement).

Distance = sum of |areas|. These two — displacement vs distance from the same graph — are the classic trap.

Δx=∫v dt=area under v–t\Delta x = \int v\, dt = \text{area under } v\text{–}t

Displacement as the v–t area

Worked example

A body's v–t graph is a triangle: 0 to 10 m/s in 5 s, then back to 0 in 5 s. Find the total distance.

  1. Distance = area under the v–t curve.
  2. First triangle: ½ × 5 × 10 = 25 m; second triangle: ½ × 5 × 10 = 25 m.
  3. Total = 50 m.

Answer: 50 m.

Transcript (0 min)
WEBVTT

1
00:00:00.000 --> 00:00:05.000
Velocity–time graphs — FemtoLearn.
2
00:00:05.000 --> 00:00:15.000
v–t graph: slope = acceleration; area under the curve = displacement.
Printable notes (free)

Practice

Free · 4 questions with full solutions
  1. Q1 · Numerical · difficulty 2/5

    The acceleration of a body starting from rest is a = 2 m/s² for 4 s, then zero for 2 s. Find its velocity at t = 6 s.

  2. Q2 · MCQ · difficulty 2/5

    A particle moves with constant velocity. Which a–t graph describes it?

    • A horizontal line at a = 0
    • A horizontal line at a = 2 m/s²
    • A straight line through the origin
    • A parabola
  3. Q3 · Numerical · difficulty 2/5

    A body moves with v = 10 m/s for 3 s, then v = 5 m/s for 2 s in the same direction. Find the total distance.

  4. Q4 · MCQ · difficulty 2/5

    The slope of a velocity–time graph gives:

    • Displacement
    • Acceleration
    • Speed
    • Distance