---
title: "Acceleration proportional to time; distance travelled"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:acceleration-proportional-to-time-distance-travelled"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/acceleration-proportional-to-time-distance-travelled
site: BCT Engineering
---

# Acceleration proportional to time; distance travelled

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

- Asked in: 2067 Asar, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/acceleration-proportional-to-time-distance-travelled
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

The acceleration of a particle is directly proportional to the time (t). At time (t) = 0, the velocity of the particle is v = 16 m/sec. Knowing that velocity (v) = 15 m/sec, position (x) = 20 m at time (t) = 1 sec, determine the velocity, the position and total distance travelled when time (t) = 7 sec.

## Short answer

Acceleration a = –2t; integrating gives v = 16 – t², x = 4.33 + 16t – t³/3. At t=7 s: v = –33 m/s, x = 2.00 m. The particle reverses at t=4 s; total distance travelled = 42.67 + 45.00 = 87.67 m.

## Answer

#### Equations
a ∝ t, so a = kt.
- v = v₀ + ∫kt dt = 16 + kt²/2
- At t = 1 s, v = 15 m/s: 15 = 16 + k/2, so **k = −2 m/s³** and a = −2t.
- v = 16 − t² (m/s)
- x = x₀ + 16t − t³/3. At t = 1 s, x = 20 m: 20 = x₀ + 16 − 1/3, so **x₀ = 4.333 m**.
- x = 4.333 + 16t − t³/3 (m)

#### At t = 7 s
- Velocity: v = 16 − 49 = **−33 m/s** (the particle moves in the negative direction).
- Position: x = 4.333 + 112 − 343/3 = 4.333 + 112 − 114.333 = **2.00 m**

#### Total distance travelled
The velocity is zero when 16 − t² = 0, so t = 4 s, and the direction of motion reverses.
- At t = 4 s: x = 4.333 + 64 − 21.333 = 47.00 m
- From t = 0 to 4 s: distance = 47.00 − 4.333 = 42.67 m
- From t = 4 to 7 s: distance = 47.00 − 2.00 = 45.00 m

**Total distance = 42.67 + 45.00 = 87.67 m** (the displacement is only 2.00 − 4.33 = −2.33 m).

**Answer: v = −33 m/s, x = 2.0 m, total distance = 87.67 m.**

## Check yourself

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

1. Given a = kt, v(0) = 16 and v(1) = 15, what is k?
   A. -2 m/s^3
   B. -1 m/s^3
   C. 2 m/s^3
   D. -0.5 m/s^3
   Answer: A. 15 = 16 + k/2 gives k = -2. (Key point: Constant k)

2. How is x0 found?
   A. Assume x0 = 0
   B. Use x(1) = 20 in x = x0 + 16t - t^3/3, giving 4.333 m
   C. Use v(1) = 15
   D. Use a = 0
   Answer: B. 20 = x0 + 16 - 1/3. (Key point: Initial position)

3. Why must distance be computed in two parts?
   A. v = 16 - t^2 is zero at t = 4 s, where the direction reverses
   B. Acceleration changes
   C. x is negative
   D. The question demands it
   Answer: A. Position at t = 4 is 47 m; going back to 2 m adds 45 m. (Key point: Direction reversal)

4. What is the total distance travelled in 7 s?
   A. 2.33 m
   B. 42.67 m
   C. 87.67 m
   D. 45 m
   Answer: C. 42.67 + 45.00 = 87.67 m; displacement is only -2.33 m. (Key point: Total distance)

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