Particle with a = v³; projectile motion equations

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°)