Consider a small particle moving in air governed by the foll

Consider a small particle moving in air, governed by the following equation: m dv/dt = -kv - mg dy/dt = v where m = 10^-4 kilograms and k = 0.1 kg/s. At t = 0 we hold it at y = 0 and let go. It starts to fall, and experiences both viscous drag and a gravitational force. What are the initial values of y, v, dy/dt, and dv/dt? Give me numbers in SI units. Now, consider the approximate definition of a derivative: f(t + Delta t) = f(t) + Delta t df/dt where df/dt is evaluated using numbers at time t, not t + Delta t. What are the values of y, v, dy/dt, and dv/dt at Delta t = 5 * 10^-5 seconds? Give me numbers in SI units. Find the values of y, v, dy/dt, and dv/dt at 2 Delta t = 2*5*10^-5 seconds.

Solution

(1) Intial values of y is 0 (no initial displacement) , v = 0 ( no initial velocity) , dy/dt = 0 , dv/dt = -g = -9.8m/s²

(2) when deltat = 5*10^-5 seconds y(0 + deltat) = y(0) + deltat(dy/dt)_t=0 => y(5*10^-5 ) = 0 + 5*10^-5 *v_t=0 = 0
   v(0 + deltat) = v(0) + deltat(dv/dt)_t=0 => v(5*10^-5 ) = 0 + 5*10^-5 *(dv/dt)_t=0 = -4.9*10^-4

dy/dt(0 + deltat) = dy/dt(0) + deltat*d²y/dt² = 0 + 5*10^-5*-9.8 = -4.9*10^-4

dv/dt(0 + deltat) = dv/dt(0) + deltat*d²v/dt² = -9.8 + 5*10^-5*((-k/m)dv/dt)_t=0 = -9.31

(3)

when deltat = 2*5*10^-5 seconds y(0 + deltat) = y(0) + deltat(dy/dt)_t=0 => y(10^-4 ) = 0 + 10^-4 *v_t=0 = 0
   v(0 + deltat) = v(0) + deltat(dv/dt)_t=0 => v(10^-4 ) = 0 + 10^-4 *(dv/dt)_t=0 = -9.8*10^-4

dy/dt(0 + deltat) = dy/dt(0) + deltat*d²y/dt² = 0 + 10^-4*-9.8 = -9.8*10^-4

dv/dt(0 + deltat) = dv/dt(0) + deltat*d²v/dt² = -9.8 + 10^-4*((-k/m)dv/dt)_t=0 = -8.82