---
title: "Particle under F = 12t i − 24t² j − 40t³ k"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:force-particle-ax-vy-impulse-momentum"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-particle-ax-vy-impulse-momentum
site: BCT Engineering
---

# Particle under F = 12t i − 24t² j − 40t³ k

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2074 Asoj, 8+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-particle-ax-vy-impulse-momentum
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

The resultant external force acting on a 2 kg particle in space is F = (12t i − 24t² j − 40t³ k) N, where t is the time measured in seconds. The particle is at rest at the origin when t = 0. Determine the acceleration component ay, the velocity component Vy, and the coordinate y of the particle at the instant of 4 sec. What do you mean by principle of impulse and momentum?

## Short answer

a_y = –12t², so at t=4 s, a_y = –192 m/s². Integrating from rest: v_y = –4t³ = –256 m/s, y = –t⁴ = –256 m. Principle of impulse and momentum: the impulse equals change in linear momentum, ∫F dt = mΔv.

## Answer

#### Equations of motion (m = 2 kg)
F = 12t i − 24t² j − 40t³ k (N). From F = ma:
- a_y = F_y/m = −24t²/2 = **−12t² m/s²**
- At t = 4 s: **a_y = −12(16) = −192 m/s²**

#### Velocity (starts from rest)
v_y = ∫a_y dt = −4t³
At t = 4 s: **v_y = −4(64) = −256 m/s**

#### Coordinate (starts at the origin)
y = ∫v_y dt = −t⁴
At t = 4 s: **y = −256 m**

(The negative signs mean that the particle moves in the −y direction.)

#### Principle of impulse and momentum
Starting from Newton's second law, F = m dv/dt, multiply by dt and integrate between times t₁ and t₂:

∫ F dt = m v₂ − m v₁

- **Linear impulse** = ∫ F dt (units N·s): the effect of a force acting over a time interval.
- **Linear momentum** = m v (units kg·m/s).
- The principle states: *the linear impulse of the resultant force on a particle during a time interval equals the change in its linear momentum.* In vector form m v₁ + ∫ F dt = m v₂, with separate component equations along x, y, z.
- If the resultant force is zero, momentum is conserved. The principle is most useful when the force is given as a function of time or when velocity and time (not displacement) are involved, for example impact.

## Check yourself

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

1. How is a_y found from the given force on the 2 kg particle?
   A. a_y = F_y x m
   B. a_y = F_y / m = -24t^2/2 = -12t^2
   C. a_y = integral of F_y dt
   D. a_y = F_y / t
   Answer: B. Newton's second law gives a_y = F_y/m, so a_y = -12t^2 m/s^2. (Key point: Acceleration from force)

2. How are v_y and y obtained from a_y, given the particle starts from rest at the origin?
   A. Differentiate a_y to get v_y and again to get y
   B. Integrate a_y to get v_y, then integrate v_y to get y, with zero constants
   C. Multiply a_y by t and t^2
   D. Use v = u + at with a = -192 for all time
   Answer: B. v_y = integral of a_y dt = -4t^3 and y = integral of v_y dt = -t^4, with zero initial values. At t = 4 s these give -256 m/s and -256 m. (Key point: Integrate for v and y)

3. What does the principle of impulse and momentum state?
   A. Impulse of the resultant force equals the change in linear momentum
   B. Work done equals the change in momentum
   C. Impulse equals the final momentum only
   D. Impulse equals force times displacement
   Answer: A. Integrating F = m dv/dt gives integral of F dt = m v2 - m v1: impulse equals the change in momentum. (Key point: Impulse-momentum principle)

## Related in Applied Mechanics

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