---
title: "Particle with a = v³; projectile motion equations"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:acceleration-a-v3-displacement-projectile-equations"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/acceleration-a-v3-displacement-projectile-equations
site: BCT Engineering
---

# Particle with a = v³; projectile motion equations

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

- Asked in: 2070 Chaitra, 7+3 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-a-v3-displacement-projectile-equations
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

The acceleration of a particle is given by a relation a = v³. It is known that at time t = 0, position is −2 m and velocity is 2 m/sec. Find the displacement, position, velocity and acceleration at instant of ½ sec. What do you mean by projectile and obtain the equations for projectile motion.

## Short answer

With a = -v³, at t=0.5 s: v=0.894 m/s, a=-0.716 m/s², position -1.382 m, displacement +0.618 m. A projectile moves under gravity alone; its equations: x = v₀ cosα t, y = v₀ sinα t - ½gt², path y = x tanα - gx²/(2v₀² cos²α).

## Answer

#### Motion with a = f(v)
The acceleration is given as a = v³, with x₀ = −2 m and v₀ = 2 m/s at t = 0. Using dv/dt = a:

dv/v³ = dt, so −1/(2v²) + 1/(2v₀²) = t, which gives 1/v² = 1/v₀² − 2t = 0.25 − 2t.

Taken literally (a = +v³) the speed becomes infinite at t = 0.125 s, so no motion exists at t = ½ s. The usual form of this problem is a **retardation a = −v³**, for which 1/v² = 0.25 + 2t. This is solved below.

- At t = 0.5 s: 1/v² = 0.25 + 1 = 1.25, so **v = 0.894 m/s**
- Acceleration: a = −v³ = −(0.894)³ = **−0.716 m/s²**
- Position: since v = (0.25 + 2t)^(−1/2), x = x₀ + ∫v dt = −2 + [(0.25 + 2t)^(1/2)] from 0 to 0.5 = −2 + (1.118 − 0.5) = **−1.382 m**
- Displacement in 0.5 s: **+0.618 m**

#### Projectile
A **projectile** is a body thrown into the air that moves under gravity alone (air resistance neglected), so the horizontal acceleration is zero and the vertical acceleration is −g. With initial speed v₀ at angle α above the horizontal, origin at the point of projection:

- ẍ = 0: vₓ = v₀ cos α, x = (v₀ cos α) t
- ÿ = −g: v_y = v₀ sin α − gt, y = (v₀ sin α) t − ½gt²
- Path: y = x tan α − g x²/(2v₀² cos²α) (a parabola)
- Time of flight (level ground): T = 2v₀ sin α/g
- Maximum height: H = v₀² sin²α/(2g)
- Range: R = v₀² sin 2α/g (maximum at α = 45°)

## Check yourself

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

1. Why does a = +v^3 fail at t = 0.5 s?
   A. 1/v^2 = 0.25 - 2t becomes zero at t = 0.125 s, so speed is infinite before 0.5 s
   B. v is always zero
   C. The position is negative
   D. Acceleration is constant
   Answer: A. Integrating dv/v^3 = dt gives 1/v^2 = 0.25 - 2t; v blows up at 0.125 s. (Key point: Literal a = +v^3)

2. For retardation a = -v^3, what is v at 0.5 s?
   A. 0.894 m/s
   B. 1.118 m/s
   C. 2 m/s
   D. 0.5 m/s
   Answer: A. 1/v^2 = 0.25 + 1 = 1.25, so v = 0.894 m/s. (Key point: Speed at 0.5 s)

3. What are the components of acceleration of a projectile?
   A. Both zero
   B. Horizontal g, vertical zero
   C. Horizontal zero, vertical -g
   D. Both -g
   Answer: C. Only gravity acts: ẍ = 0, ÿ = -g. (Key point: Horizontal acceleration)

4. For which angle is the range v0^2 sin 2α/g maximum on level ground?
   A. 30 deg
   B. 90 deg
   C. 60 deg
   D. 45 deg
   Answer: D. sin 2α is maximum when 2α = 90 deg. (Key point: Maximum range)

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