---
title: "Define kinematic quantities; a = −2 m/s² distance"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:position-velocity-acceleration-total-distance"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/position-velocity-acceleration-total-distance
site: BCT Engineering
---

# Define kinematic quantities; a = −2 m/s² distance

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

- Asked in: 2069 Asar, 3+7 marks
- Video: [Rectilinear Motion Problem No.3 - Kinematics of Particles - Engineering Mechanics](https://www.youtube.com/watch?v=P1_V8GhSLsE) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/position-velocity-acceleration-total-distance
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

Define position, velocity and acceleration of a particle. The acceleration of a particle is defined by the relation a = −2 m/s². Initially if the velocity of the particle is 10 m/s at 0 m position, determine the velocity, position and the total distance travelled at the instant of 5 seconds.

## Short answer

Position is location, velocity is rate of change of position, acceleration is rate of change of velocity. With a = -2 m/s², v₀=10 m/s, at t=5 s: v=0 m/s, position 25 m, total distance 25 m (motion in one direction).

## Answer

#### Definitions
- **Position (x):** the location of the particle on its path, measured from a chosen origin (m).
- **Velocity (v):** the rate of change of position, v = dx/dt (m/s). Average velocity is Δx/Δt.
- **Acceleration (a):** the rate of change of velocity, a = dv/dt = v dv/dx (m/s²).

#### Problem
a = −2 m/s², v₀ = 10 m/s, x₀ = 0.
- v = v₀ + at = 10 − 2t
- x = 10t − t²

**At t = 5 s:**
- v = 10 − 2(5) = **0 m/s**
- x = 10(5) − 25 = **25 m**

**Total distance:** the velocity is zero at t = 10/2 = 5 s and positive before that, so the particle moves in one direction only up to t = 5 s. The distance travelled equals the displacement:

**Total distance = 25 m.** (If the motion is continued beyond 5 s, the particle returns; at 5 s it has just stopped.)

## Check yourself

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

1. Which is correct?
   A. v = dx/dt, a = dv/dt
   B. v = dv/dt, a = dx/dt
   C. a = dx/dt
   D. v = x/a
   Answer: A. Velocity is the rate of change of position; acceleration is the rate of change of velocity. (Key point: Definitions)

2. With a = -2, v0 = 10, x0 = 0, what are v and x at 5 s?
   A. v = 0, x = 25 m
   B. v = 10, x = 50 m
   C. v = -10, x = 0
   D. v = 0, x = 50 m
   Answer: A. v = 10 - 10 = 0 and x = 50 - 25 = 25 m. (Key point: State at 5 s)

3. Why is total distance equal to displacement here?
   A. The velocity never reverses before 5 s (v reaches zero exactly at 5 s)
   B. Acceleration is zero
   C. The particle returns
   D. Distance is always equal to displacement
   Answer: A. v = 10 - 2t is positive until 5 s, so there is no reversal. (Key point: Total distance)

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