---
title: "Particle with x = t³/3 − 6t² − 15t"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:particle-x-t3-3-6t2-15t-velocity-zero"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/particle-x-t3-3-6t2-15t-velocity-zero
site: BCT Engineering
---

# Particle with x = t³/3 − 6t² − 15t

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

- Asked in: 2068 Shrawan
- Video: [Dynamics - Position of a Particle (When is the velocity zero and total distanced traveled)](https://www.youtube.com/watch?v=nbn2-S09DaM) — Engineering Corner
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/particle-x-t3-3-6t2-15t-velocity-zero
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

The position of a particle which moves along a straight line is defined by the relation x = t³/3 − 6t² − 15t, where x is in metre and t in seconds. Determine: i) The time at which velocity will be zero. ii) The position and distance travelled by the particle at that time. iii) The acceleration of the particle at that time.

## Short answer

Velocity zero at t=13.14 s. Position -476.8 m, distance 476.8 m, acceleration 14.28 m/s².

## Answer

#### Velocity and acceleration
x = t³/3 − 6t² − 15t
- v = t² − 12t − 15 (m/s)
- a = 2t − 12 (m/s²)

**(i) Time when v = 0:** t² − 12t − 15 = 0, so t = 6 + √51 = **13.14 s** (the negative root is rejected).

**(ii) Position and distance travelled at that time**
x = 13.14³/3 − 6(13.14)² − 15(13.14) = 756.5 − 1036.1 − 197.1 = **−476.8 m**
At t = 0, v = −15 m/s. The velocity first becomes more negative, reaching its minimum of −51 m/s at t = 6 s (where a = 0), and then rises to zero at 13.14 s. It stays negative throughout, so the particle moves only in the negative direction and the distance travelled is **476.8 m**.

**(iii) Acceleration at that time:** a = 2(13.14) − 12 = **14.28 m/s²**

## Check yourself

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

1. Which is the time at which v = 0?
   A. t = 6 + sqrt(51) = 13.14 s from t^2 - 12t - 15 = 0
   B. t = 6 s
   C. t = 15 s
   D. t = 6 - sqrt(51)
   Answer: A. v = t^2 - 12t - 15 and the negative root is rejected. (Key point: Time of zero velocity)

2. Why is the distance 476.8 m (equal to |x|)?
   A. v < 0 for the whole interval, so the particle never reverses
   B. v > 0 throughout
   C. a = 0
   D. The distance is the sum of two intervals
   Answer: A. v starts at -15 m/s and increases to zero, so motion is in one direction. (Key point: Distance travelled)

3. What is a at that time?
   A. 12 m/s^2
   B. 14.28 m/s^2
   C. 0
   D. 26.28 m/s^2
   Answer: B. a = 2t - 12 = 26.28 - 12 = 14.28 m/s^2. (Key point: Acceleration)

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