An object of mass 0.14 kg is released from rest. Take downward as positive and model drag by F_drag = −bv with b = 0.05 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 = 2.8 s; terminal speed 27.47 m s⁻¹; v(2τ) = 23.75 m s⁻¹; distance 87.32 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 = 2.8 s and v_t = 27.47 m s⁻¹. At t = 2τ the remaining fractional difference from terminal speed is exp(−2). Hence v(2τ) = 23.75 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)) = 87.32 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 2

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