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

# Velocity and acceleration of particle on curve

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2073 Bhadra, 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-particle-curve
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

A particle moves along the curve x = a cos t, y = a sin t and z = bt. Find the velocity and acceleration at t = 0 and t = π/2.

## Short answer

Differentiate r: v = -a sin t i + a cos t j + b k, f = -a cos t i - a sin t j. At t=0: v = a j + b k, f = -a i. At t=π/2: v = -a i + b k, f = -a j.

## Answer

The position vector is r = a cos t i + a sin t j + bt k (a circular helix). Velocity v = dr/dt and acceleration f = d²r/dt².

#### Velocity
```text
v = dr/dt = −a sin t i + a cos t j + b k
```

#### Acceleration
```text
f = d²r/dt² = −a cos t i − a sin t j
```

#### At t = 0
```text
v = −a(0) i + a(1) j + b k = a j + b k
f = −a i
```
Speed = |v| = √(a² + b²).

#### At t = π/2
```text
v = −a(1) i + a(0) j + b k = −a i + b k
f = −a(0) i − a(1) j = −a j
```
Speed = |v| = √(a² + b²).

**Answer:**
- t = 0: velocity = a j + b k, acceleration = −a i.
- t = π/2: velocity = −a i + b k, acceleration = −a j.

The speed √(a² + b²) is constant, and the acceleration has constant magnitude a, directed towards the z-axis.

## Check yourself

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

1. Velocity and acceleration are:
   A. v = d²r/dt², f = dr/dt
   B. v = r/t
   C. v = ∫r dt
   D. v = dr/dt, f = d²r/dt²
   Answer: D. Derivatives of position. (Key point: Definitions)

2. v for r = a cos t i + a sin t j + bt k is:
   A. a sin t i − a cos t j + b k
   B. −a sin t i + a cos t j
   C. −a sin t i + a cos t j + b k
   D. −a cos t i − a sin t j
   Answer: C. Differentiate each component. (Key point: Velocity)

3. At t = 0 the acceleration is:
   A. a j
   B. −a i
   C. −a j
   D. 0
   Answer: B. f = −a cos t i − a sin t j. (Key point: At t = 0)

4. The speed |v| is:
   A. √(a² + b²), constant
   B. a, constant
   C. √(a² + b²) sin t
   D. b
   Answer: A. The a sin/cos terms cancel in the magnitude. (Key point: Speed)

## Close variants set in other exams

- [Velocity and acceleration on curve x=4cos t, y=t², z=2t](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Avelocity-acceleration-curve-x-4cos) — Mathematics, chapter 5, asked 2×

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