---
title: "Velocity and acceleration of a particle"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:velocity-and-acceleration-of-a-particle"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/velocity-and-acceleration-of-a-particle
site: BCT Engineering
---

# Velocity and acceleration of a particle

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2067 Asar
- Video: [Velocity and Position From Acceleration By Integration - Physics and Calculus](https://www.youtube.com/watch?v=UNNXPxekbl8) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/velocity-and-acceleration-of-a-particle
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

The acceleration of a moving particle at any time t is given by d²r/dt² = 12 cos 2t i − 8 sin 2t j + 16t k. Find the velocity v and displacement r at any time t if t = 0, v = 0 and r = 0.

## Short answer

Integrate acceleration twice with initial conditions v(0)=0, r(0)=0. v = 6 sin 2t i + 4(cos 2t − 1) j + 8t² k, r = 3(1 − cos 2t) i + (2 sin 2t − 4t) j + (8t³/3) k.

## Answer

Given d²r/dt² = 12cos 2t i − 8sin 2t j + 16t k, with v = 0 and r = 0 at t = 0. Integrate twice and use the initial conditions.

#### Velocity
```text
v = dr/dt = ∫ (12cos 2t i − 8sin 2t j + 16t k) dt
  = 6 sin 2t i + 4 cos 2t j + 8t² k + c₁
At t = 0, v = 0:  0 + 4 j + 0 + c₁ = 0  → c₁ = −4 j
v = 6 sin 2t i + (4 cos 2t − 4) j + 8t² k
```

#### Displacement
```text
r = ∫ v dt
  = −3 cos 2t i + (2 sin 2t − 4t) j + (8t³/3) k + c₂
At t = 0, r = 0:  −3 i + c₂ = 0  → c₂ = 3 i
r = 3(1 − cos 2t) i + (2 sin 2t − 4t) j + (8t³/3) k
```

**Answer:**
- v = 6 sin 2t i + 4(cos 2t − 1) j + 8t² k
- r = 3(1 − cos 2t) i + (2 sin 2t − 4t) j + (8t³/3) k

## Check yourself

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

1. How are v and r obtained from the acceleration?
   A. Differentiate twice
   B. Integrate once and set r = vt
   C. Multiply a by t^2/2 only
   D. Integrate twice, fixing each constant with the initial conditions
   Answer: D. v = integral of a, r = integral of v, using v(0)=0 and r(0)=0 for the constants. (Key point: Method)

2. Why is c1 = -4 j?
   A. At t=0 the integral gives 4j, which must cancel to give v = 0
   B. Because sin 2t is 0 at t = 0
   C. Because the k term is 8t^2
   D. Because r(0) = 0
   Answer: A. 4cos(0) j = 4 j, so c1 = -4 j. (Key point: Velocity constant)

3. Which is v?
   A. 6 sin2t i + 4cos2t j + 8t^2 k
   B. 6 sin2t i + 4(cos2t - 1) j + 8t^2 k
   C. -6 sin2t i + 4(cos2t - 1) j + 8t^2 k
   D. 12 sin2t i + 4(cos2t - 1) j + 16t^2 k
   Answer: B. The integral of 12cos2t is 6 sin2t, of -8sin2t is 4cos2t, of 16t is 8t^2. (Key point: Velocity result)

4. What is the i-component of r?
   A. -3 cos 2t
   B. 3 cos 2t
   C. 3(1 - cos 2t)
   D. 3(1 + cos 2t)
   Answer: C. Integrating 6 sin2t gives -3cos2t; r(0)=0 gives c2 = 3, so 3(1 - cos2t). (Key point: Displacement)

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