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

# Velocity and acceleration of a particle

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

- Asked in: 2067 Magh
- Video: [Velocity Acceleration Problem 1](https://www.youtube.com/watch?v=4kOPwGeuV2k) — TutorialsPoint
- 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-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

The position vector of a moving particle at any time t is given by r = (t² + 1) i + (4t − 3) j + (2t² − 6) k. Find the velocity and acceleration at t = 1. Also find their magnitudes.

## Short answer

Differentiate: v = dr/dt = 2t i + 4 j + 4t k, a = dv/dt = 2 i + 4 k. At t=1, v = 2i+4j+4k, magnitude |v| = 6; a = 2i+4k, magnitude |a| = 2√5 ≈ 4.47.

## Answer

Velocity v = dr/dt and acceleration a = d²r/dt².

```text
r = (t² + 1) i + (4t − 3) j + (2t² − 6) k
v = dr/dt = 2t i + 4 j + 4t k
a = dv/dt = 2 i + 0 j + 4 k
```

At t = 1:

```text
v = 2 i + 4 j + 4 k
a = 2 i + 4 k
|v| = √(4 + 16 + 16) = √36 = 6
|a| = √(4 + 0 + 16) = √20 = 2√5 ≈ 4.47
```

**Answer:** v(1) = 2i + 4j + 4k with |v| = 6; a(1) = 2i + 4k with |a| = 2√5 ≈ 4.47.

## Check yourself

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

1. How is velocity found from position?
   A. v = integral of r dt
   B. v = d^2r/dt^2
   C. v = dr/dt
   D. v = r/t
   Answer: C. Velocity is the first derivative. (Key point: Velocity)

2. What is a at any t?
   A. 2t i + 4 j + 4t k
   B. 2 i + 4 j + 4 k
   C. 4 k
   D. 2 i + 4 k
   Answer: D. Differentiate v = 2t i + 4j + 4t k. (Key point: Acceleration)

3. |v| at t = 1 is:
   A. 6
   B. sqrt(20)
   C. 36
   D. 4
   Answer: A. v = 2i+4j+4k, |v| = sqrt(36) = 6. (Key point: Magnitude)

4. |a| equals:
   A. 6
   B. 2 sqrt(5)
   C. sqrt(24)
   D. 4
   Answer: B. a = 2i + 4k, |a| = sqrt(20) = 2 sqrt 5. (Key point: Magnitude of a)

## Related in Mathematics III

- [Velocity and acceleration of a particle](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Avelocity-and-acceleration-of-a-particle) — Mathematics III, chapter 2

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