---
title: "Velocity and acceleration of moving particle at t=1"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:velocity-acceleration-moving-particle-t"
source: https://www.bctengineer.com/q/semester_2/mathematics/velocity-acceleration-moving-particle-t
site: BCT Engineering
---

# Velocity and acceleration of moving particle at t=1

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2067 Chaitra, 5 marks
- Video: [Multivariable Calculus: Velocity and Acceleration](https://www.youtube.com/watch?v=CM04hZW7WL4) — Dr. Bevin Maultsby
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/velocity-acceleration-moving-particle-t
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

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

## Short answer

Velocity v = 2i + 4j + 4k, magnitude 6. Acceleration a = 2i + 0j + 4k, magnitude 2√5 ≈ 4.47.

## Answer

r = (t² + 1) i + (4t − 3) j + (2t² − 6) k.

#### Velocity and acceleration
```text
v = dr/dt = 2t i + 4 j + 4t k
f = d²r/dt² = 2 i + 0 j + 4 k
```

#### At t = 1
```text
v = 2i + 4j + 4k
f = 2i + 4k
```

#### Magnitudes
```text
|v| = √(4 + 16 + 16) = √36 = 6
|f| = √(4 + 0 + 16) = √20 = 2√5 ≈ 4.47
```

**Answer: at t = 1, velocity = 2i + 4j + 4k with magnitude 6; acceleration = 2i + 4k with magnitude 2√5 ≈ 4.47.**

## Check yourself

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

1. For r = (t²+1)i + (4t−3)j + (2t²−6)k, what is v = dr/dt?
   A. 2t i + 4 j + 2t k
   B. 2t i + 4 j + 4t k
   C. (t²/3 + t) i + ...
   D. 2 i + 0 j + 4 k
   Answer: B. Differentiate each component: d(t²+1)/dt = 2t, d(4t−3)/dt = 4, d(2t²−6)/dt = 4t. (Key point: Velocity from position)

2. What is the acceleration f = d²r/dt² of the particle?
   A. 2t i + 4 j + 4t k
   B. 2 i + 4 j + 4 k
   C. 2 i + 4 k
   D. 4 j
   Answer: C. Differentiate v again: 2t gives 2, 4 gives 0, 4t gives 4, so f = 2i + 4k, constant. (Key point: Acceleration)

3. At t = 1, the velocity is 2i + 4j + 4k. What are |v| and |f|?
   A. |v| = 10, |f| = 6
   B. |v| = 6, |f| = 2√5
   C. |v| = 6, |f| = 6
   D. |v| = 2√5, |f| = 6
   Answer: B. |v| = √(4+16+16) = 6 and |f| = √(4+0+16) = √20 = 2√5 ≈ 4.47. (Key point: Magnitude at t = 1)

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