---
title: "Velocity and acceleration on curve x=4cos t, y=t², z=2t"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:velocity-acceleration-curve-x-4cos"
source: https://www.bctengineer.com/q/semester_2/mathematics/velocity-acceleration-curve-x-4cos
site: BCT Engineering
---

# Velocity and acceleration on curve x=4cos t, y=t², z=2t

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2070 Bhadra, 5 marks
- Video: [Motion on a Curve (How to find velocity and acceleration vector function) 9.2   #engineering](https://www.youtube.com/watch?v=oT07cU_gEdA) — Lakhani STEM Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/velocity-acceleration-curve-x-4cos
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

A particle moves along the curve x = 4 cos t, y = t², z = 2t. Find velocity and acceleration at time t = 0 and t = π/2.

## Short answer

Differentiate r = 4 cos t i + t² j + 2t k to get v = −4 sin t i + 2t j + 2k and a = −4 cos t i + 2 j. At t=0, v = 2k, a = −4i+2j. At t=π/2, v = −4i+πj+2k, a = 2j.

## Answer

Position: r = 4 cos t i + t² j + 2t k. Velocity v = dr/dt, acceleration f = d²r/dt².

#### Velocity and acceleration
```text
v = −4 sin t i + 2t j + 2 k
f = −4 cos t i + 2 j
```

#### At t = 0
```text
v = −4(0) i + 0 j + 2 k = 2k
f = −4(1) i + 2 j = −4i + 2j
```

#### At t = π/2
```text
v = −4(1) i + π j + 2 k = −4i + πj + 2k
f = −4(0) i + 2 j = 2j
```
(Speed at t = π/2: √(16 + π² + 4) = √(20 + π²) ≈ 5.46.)

**Answer:**
- t = 0: velocity = 2k, acceleration = −4i + 2j.
- t = π/2: velocity = −4i + πj + 2k, acceleration = 2j.

## Check yourself

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

1. v for r = 4cos t i + t² j + 2t k is:
   A. 4 sin t i + 2t j + 2k
   B. −4 sin t i + 2t j + 2k
   C. −4 sin t i + t j + 2k
   D. −4 cos t i + 2 j
   Answer: B. Differentiate each term. (Key point: Velocity)

2. f is:
   A. −4 cos t i + 2 j
   B. −4 sin t i + 2 j
   C. −4 cos t i + 2t j
   D. −4 cos t i + 2 j + 2k
   Answer: A. Differentiate v. (Key point: Acceleration)

3. At t = 0:
   A. v = 2k, f = 2j
   B. v = −4i + 2k, f = −4i
   C. v = 0, f = 0
   D. v = 2k, f = −4i + 2j
   Answer: D. sin 0 = 0 and cos 0 = 1. (Key point: At t = 0)

4. At t = π/2 velocity is:
   A. −4i + 2j + 2k
   B. πj + 2k
   C. −4i + πj + 2k
   D. 4i + πj + 2k
   Answer: C. 2t = π. (Key point: At t = π/2)

## Close variants set in other exams

- [Velocity and acceleration of particle on curve](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Avelocity-acceleration-particle-curve) — Mathematics, chapter 5, asked 2×

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- [Velocity and acceleration of moving particle at t=1](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Avelocity-acceleration-moving-particle-t) — Mathematics, chapter 5

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