An object of mass 0.24 kg is released from rest. Take downward as positive and model drag by F_drag = −bv with b = 0.075 kg s⁻¹. Ignore buoyancy and use g = 9.81 m s⁻². (a) Solve m dv/dt = mg − bv for v(t). (b) Calculate terminal speed, the time constant τ, speed at 2τ, and distance fallen by 2τ.

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v(t) = (mg/b)[1 − exp(−t/τ)], τ = 3.2 s; terminal speed 31.39 m s⁻¹; v(2τ) = 27.14 m s⁻¹; distance 114 m. Separate the differential equation or use an integrating factor. Its steady solution is v_t = mg/b, and the decaying homogeneous term is A exp(−bt/m). The initial condition v(0) = 0 fixes A = −v_t. Thus v(t) = v_t[1 − exp(−t/τ)], with τ = m/b = 3.2 s and v_t = 31.39 m s⁻¹. At t = 2τ the remaining fractional difference from terminal speed is exp(−2). Hence v(2τ) = 27.14 m s⁻¹, less than v_t. Integrating velocity from 0 to t gives y(t) = v_t[t − τ(1 − exp(−t/τ))]. At 2τ this becomes v_tτ(1 + exp(−2)) = 114 m. Differentiating the velocity gives the initial acceleration g and an acceleration approaching zero at long times. The short-time expansion yields y ≈ ½gt², which independently checks the integration constant and sign.

A falling object with linear drag · set 7

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