---
title: "Motion when acceleration is function of time"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:motion-when-acceleration-function-of-time"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/motion-when-acceleration-function-of-time
site: BCT Engineering
---

# Motion when acceleration is function of time

Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body

- Asked in: 2068 Shrawan
- Video: [Example 1D Motion Acceleration As A Function of Time - Engineering Dynamics](https://www.youtube.com/watch?v=4KHNJooddU8) — structurefree
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/motion-when-acceleration-function-of-time
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

How is the motion of a particle found when the acceleration is a given function of time?

## Short answer

Integrate acceleration with respect to time to get velocity: v = v₀ + ∫ a dt. Then integrate velocity to get position: x = x₀ + ∫ v dt. Use initial conditions v₀, x₀.

## Answer

#### Motion when a is a given function of time
If a = f(t), velocity and position are found by successive integration, using the initial conditions.

1. Velocity: a = dv/dt, so dv = f(t) dt, and
 v = v₀ + ∫₀ᵗ f(t) dt
2. Position: v = dx/dt, so
 x = x₀ + ∫₀ᵗ v dt
3. The constants v₀ and x₀ are the velocity and position at t = 0 (or any known instant).

Example: a = 6t m/s² with v₀ = 2 m/s and x₀ = 0:
- v = 2 + ∫6t dt = 2 + 3t²
- x = ∫(2 + 3t²) dt = 2t + t³

At t = 2 s: v = 14 m/s and x = 12 m.

If the particle changes direction (v passes through zero), the total distance travelled is found by adding the distances in each interval separately.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How are v and x found from a = f(t)?
   A. Differentiate a twice
   B. Integrate: v = v0 + integral of f dt, then x = x0 + integral of v dt
   C. Multiply a by t
   D. Use v^2 = u^2 + 2as
   Answer: B. Successive integration with initial conditions. (Key point: Method)

2. What are the constants of integration?
   A. Always zero
   B. The velocity and position at t = 0 or another known instant
   C. The maximum acceleration
   D. The time interval
   Answer: B. v0 and x0 come from initial conditions. (Key point: Constants)

3. If velocity passes through zero, how is total distance found?
   A. It equals the displacement
   B. Add the distances of each interval separately
   C. Use only the last interval
   D. Multiply by two
   Answer: B. Displacement cancels the back-travel, so distances must be added piece by piece. (Key point: Direction reversal)

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