Question

The rate of change of the downward velocity of a falling object is the acceleration of gravity (10 meters/sec^2) minus the acceleration due to air resistance. Suppose that the acceleration due to air resistance is 0.25 inverse seconds times the downward velocity. Write the initial value problem and the solution for the downward velocity v for an object that is dropped (not thrown) from a great height.

dv/dt=
v(0)=
v(t)=

Answer 1

Answer:

dv / dt = 9.81 - 0.25*v

v(0) = 0

v (t) = 39.24 * ( 1 - e ^ (-t / 4 ) )

Explanation:

Given:

a_air = - 0.25 * v

a_g = 9.81 m/s^2

The rate at which the object's velocity will vary is given:

dv / dt = 9.81 - 0.25*v

Since, the ball is dropped the initial velocity is assumed to be 0, v (0) = 0

The following relation for velocity at time t is given by solving the above ODE wrt to time t.

Separating variables:

dv / (9.81 - 0.25*v) = dt

Integrating both sides

-4*Ln (9.81 - 0.25*v) = t + C

Ln (9.81 - 0.25*v) = -t / 4 + C

Evaluate C for v(0) = 0

C = Ln (9.81)

Making and explicit function of velocity v:

9.81 - 0.25*v = e ^ (-t / 4  + Ln (9.81))

0.25*v = 9.81 - 9.81*e ^ (-t / 4 )

v (t) = 39.24 * ( 1 - e ^ (-t / 4 ) )