Variable acceleration: the calculus way

When a is not constant, integrate: v = ∫a dt, x = ∫v dt — with the constants from initial conditions.

Variable acceleration: the calculus way0 min · Free lecture

In this lesson

  • When a is not constant, integrate: v = ∫a dt, x = ∫v dt — with the constants from initial conditions.

a = 6t is not constant — the SUVAT equations die here; calculus takes over.

v(t) = v₀ + ∫a dt; x(t) = x₀ + ∫v dt. When a is a function of x instead, use v dv = a dx (the chain-rule form) — a JEE favourite.

Always write the integration constants explicitly using t = 0 conditions.

vdvdx=av\frac{dv}{dx} = a

Acceleration in the v–x form

Worked example

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.

  1. Area under the a–t graph = change in velocity.
  2. Area = 2 m/s² × 4 s = 8 m/s (the last 2 s add nothing).

Answer: 8

Transcript (0 min)
WEBVTT

1
00:00:00.000 --> 00:00:05.000
Variable acceleration: the calculus way — FemtoLearn.
2
00:00:05.000 --> 00:00:15.000
When a is not constant, integrate: v = ∫a dt, x = ∫v dt — with the constants from initial conditions.
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 · 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.

  3. Q3 · Numerical · difficulty 2/5

    A body starts from rest with a = 4 m/s². Find the displacement in the 5th second.

  4. Q4 · 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