Assuming that only air resistance and gravity act on a falling object, we can find that the velocity of the object, v, must obey the differential equation dv m mg bv dt  . Here, m is the mass of the object, g is the acceleration due to gravity, and b > 0 is a constant. Consider an object that has a mass of 100 kilograms and an initial velocity of 10 m/sec (that is, v (0) = 10). If we take g to be 9.8 m/sec2 and b to be 5 kg/sec, find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

v = 196 - 186*e^ ( - 0.05*t)

v-terminal = 196 m/s

Explanation:

Given:

- The differential equation for falling object velocity v in gravity with air resistance is given by:

m*dv/dt = m*g - b*v

- The initial conditions and constants are as follows:

v (0) = 10, m = 100 kg, b = 5 kg/s, g = 9.8 m/s^2

Find:

- Find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

Solution:

- Rewrite the differential equation in te form:

dv/dt + (b/m) * v = g

- The integration factor function P (t) = b/m. The integrating factor u (t) is:

u (t) = e^∫P (t). dt

u (t) = e^∫ (b/m). dt

u (t) = e^[ (b/m). t]

- Solve the differential equation after expressing in form:

v. u (t) = ∫u (t). g. dt

v. e^[ (b/m). t] = g*∫e^[ (b/m). t]. dt

v. e^[ (b/m). t] = g*m*e^[ (b/m). t] / b + C

v = g*m/b + C*e^[ - (b/m). t]

- Apply the initial conditions v (0) = 10 m/s and evaluate C:

10 = 9.8*100/5 + C*e^[ - (b/m).0]

10 = 9.8*100/5 + C

C = - 186

- The final ODE solution is:

v = 196 - 186*e^ ( - 0.05*t)

- The Terminal velocity vt can be expressed by a limiting value for v (t), where t - >∞.

vt = Lim t - >∞ (v (t))

vt = Lim t - >∞ (196 - 186*e^ ( - 0.05*t))

vt = 196 - 0 = 196 m/s