Detailed Solution
1. For Stopping Time (τ): dv / dt = −a√v ⇒ v⁻¹/² dv = −a dt Integrating from v₀ to 0 and t = 0 to τ: [2√v] from v₀ to 0 = −aτ ⇒ −2√v₀ = −aτ ⇒ τ = (2√v₀) / a
2. For Stopping Distance (s): v (dv / dx) = −a√v ⇒ v¹/² dv = −a dx Integrating from v₀ to 0 and x = 0 to s: [(2/3)v³/²] from v₀ to 0 = −as ⇒ s = (2v₀³/²) / (3a)
Irodov 1.13 Visualizer: Square-Root Drag (w = a√v)
τ = 2√v₀ / a | sₘₐₓ = 2v₀³/² / (3a) | ⟨v⟩ = v₀ / 3
Time (t)
0.00 s
Velocity v(t)
16.00 m/s
Retardation w(t)
8.00 m/s²
Distance s(t)
0.00 m
Stop (τ | sₘₐₓ)
4.00s | 21.33m
1D RESISTIVE STOPPING TRACK (FINITE STOP vs. VISCOUS CREEP)
● w = a√v (Stops at τ)
○ Viscous Drag w = kv (Never stops)
VELOCITY DECAY v(t) = (√v₀ − ½at)²
━ v(t) Parabola ┅ ⟨v⟩ = v₀/3
DISTANCE s(t) & PHASE SPACE v(s)
━ s(t) Curve ┆ Stop Wall τ
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