---
title: "Velocity and acceleration for r = 4θ², θ = t²/2"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:polar-path-r-4theta2-velocity-acceleration"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/polar-path-r-4theta2-velocity-acceleration
site: BCT Engineering
---

# Velocity and acceleration for r = 4θ², θ = t²/2

Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body

- Asked in: 2071 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/polar-path-r-4theta2-velocity-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

A particle moves along a curved path defined by r = 4θ² and θ = t²/2 where r is in meters and t in seconds. Determine the velocity and acceleration of the particle when θ = 80°.

## Short answer

At θ=80°, compute θ̇=1.671 rad/s, r=7.798 m, ṙ=18.67 m/s. Velocity: v_r=18.67 m/s, v_θ=13.03 m/s, magnitude 22.8 m/s. Acceleration: a_r=11.73 m/s², a_θ=70.19 m/s², magnitude 71.2 m/s².

## Answer

#### Polar components
Given r = 4θ² and θ = t²/2. At θ = 80° = 1.3963 rad:
- θ̇ = t = √(2θ) = √2.7925 = 1.6711 rad/s, θ̈ = 1 rad/s²
- r = 4θ² = 4(1.3963)² = 7.798 m
- ṙ = 8θθ̇ = 8(1.3963)(1.6711) = 18.67 m/s
- r̈ = 8θ̇² + 8θθ̈ = 8(2.7925) + 8(1.3963) = 33.51 m/s²

#### Velocity
- v_r = ṙ = 18.67 m/s
- v_θ = rθ̇ = 7.798 × 1.6711 = 13.03 m/s
- v = √(18.67² + 13.03²) = **22.77 m/s**, at an angle tan⁻¹(13.03/18.67) = 34.9° from the radial line.

#### Acceleration
- a_r = r̈ − rθ̇² = 33.51 − 7.798(2.7925) = 33.51 − 21.78 = 11.73 m/s²
- a_θ = rθ̈ + 2ṙθ̇ = 7.798(1) + 2(18.67)(1.6711) = 7.80 + 62.40 = 70.19 m/s²
- a = √(11.73² + 70.19²) = **71.16 m/s²**, at 80.5° from the radial line.

**Answer: v = 22.8 m/s (v_r = 18.67 m/s, v_θ = 13.03 m/s); a = 71.2 m/s² (a_r = 11.73 m/s², a_θ = 70.19 m/s²).**

## Check yourself

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

1. Given θ = t^2/2, how is θ̇ at θ = 80 deg found?
   A. θ̇ = t = sqrt(2θ) = 1.671 rad/s (θ in radians)
   B. θ̇ = 80 deg/s
   C. θ̇ = θ/t^2
   D. θ̇ = 1 rad/s
   Answer: A. θ̇ = t and t = sqrt(2θ) with θ = 1.3963 rad. (Key point: Angular rate)

2. Which gives the polar velocity components?
   A. v_r = rθ̇, v_θ = ṙ
   B. v_r = ṙ, v_θ = rθ̇
   C. v_r = r̈, v_θ = rθ̈
   D. v_r = r, v_θ = θ
   Answer: B. Radial speed is ṙ; transverse speed is rθ̇. (Key point: Velocity components)

3. What is a_r?
   A. r̈
   B. r̈ - rθ̇^2
   C. r̈ + rθ̇^2
   D. rθ̈ + 2ṙθ̇
   Answer: B. a_r = r̈ - rθ̇^2 = 33.51 - 21.78 = 11.73 m/s^2. (Key point: Radial acceleration)

4. Which includes the Coriolis-type term?
   A. a_θ = rθ̈ + 2ṙθ̇
   B. a_θ = rθ̈
   C. a_θ = rθ̇^2
   D. a_θ = ṙθ̇
   Answer: A. a_θ = rθ̈ + 2ṙθ̇ = 7.80 + 62.40 = 70.19 m/s^2. (Key point: Transverse acceleration)

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