---
title: "Position when velocity and acceleration given"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:particle-position-x-t3-3-3t2-8t-15"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/particle-position-x-t3-3-3t2-8t-15
site: BCT Engineering
---

# Position when velocity and acceleration given

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

- Asked in: 2071 Chaitra, 6 marks
- Video: [❖ Position, Velocity, Acceleration using Derivatives ❖](https://www.youtube.com/watch?v=pFeuGMMiZWw) — Patrick J
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/particle-position-x-t3-3-3t2-8t-15
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

Motion of a particle is defined by a relation x = t³/3 − 3t² + 8t + 15. Determine the position of particle when velocity is 2.5 m/sec. Also determine the position of particle when acceleration is 3.6 m/sec².

## Short answer

When v=2.5 m/s, solving t²-6t+8=2.5 gives t=1.129 s (x=20.69 m) and t=4.871 s (x=21.31 m). When a=3.6 m/s², 2t-6=3.6 gives t=4.8 s, x=21.14 m.

## Answer

#### Velocity and acceleration
- x = t³/3 − 3t² + 8t + 15
- v = dx/dt = t² − 6t + 8 (m/s)
- a = dv/dt = 2t − 6 (m/s²)

#### Position when v = 2.5 m/s
t² − 6t + 8 = 2.5, so t² − 6t + 5.5 = 0, giving t = 3 ± √3.5 = 4.871 s or 1.129 s.
- t = 4.871 s: x = 4.871³/3 − 3(4.871)² + 8(4.871) + 15 = 38.52 − 71.18 + 38.97 + 15 = **21.31 m**
- t = 1.129 s: x = 0.480 − 3.824 + 9.033 + 15 = **20.69 m**

#### Position when a = 3.6 m/s²
2t − 6 = 3.6, so t = 4.8 s.
x = 4.8³/3 − 3(4.8)² + 8(4.8) + 15 = 36.86 − 69.12 + 38.4 + 15 = **21.14 m** (v at that instant = 4.8² − 6(4.8) + 8 = 2.24 m/s).

**Answer: x = 21.31 m or 20.69 m when v = 2.5 m/s; x = 21.14 m when a = 3.6 m/s².**

## Check yourself

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

1. For x = t^3/3 - 3t^2 + 8t + 15, what are v and a?
   A. v = t^2 - 6t + 8, a = 2t - 6
   B. v = t^3 - 6t + 8, a = 3t^2 - 6
   C. v = t^2/3 - 6t + 8, a = 2t
   D. v = 2t - 6, a = 2
   Answer: A. v = dx/dt and a = dv/dt. (Key point: Differentiation)

2. Why are there two positions when v = 2.5 m/s?
   A. t^2 - 6t + 5.5 = 0 has two positive roots (1.129 s and 4.871 s)
   B. Velocity is constant
   C. Position is symmetric
   D. Two accelerations
   Answer: A. The quadratic in t gives t = 3 ± sqrt(3.5), and each time has its own position. (Key point: Two times for v = 2.5)

3. When a = 3.6 m/s^2, what is t?
   A. t = 4.8 s
   B. t = 1.8 s
   C. t = 9.6 s
   D. t = 3.3 s
   Answer: A. 2t - 6 = 3.6 gives t = 4.8 s, then x = 21.14 m. (Key point: Acceleration condition)

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