Chapter 7.4, Problem 53E

a.

To determine

Show that the velocity of an object slowed by air resistance satisfies the differential equation.

  $m\frac{dy}{dt}=-kv$ .

Explanation of Solution

Given information: The resistance force on an object of mass m moving with velocity v is thus −kv for some positive constant k.

Proof:

Acceleration is the instantaneous rate of change of velocity so acceleration = dv/dt.

If m is the mass, then Force = mass x acceleration $=m\frac{dv}{dt}$ .

It is also give that Force = -kv .

Therefore, $m\frac{dv}{dt}=force=-kv$ .

b.

To determine

To solve the differential equation to show that $v=v_0{e}^{-\left(k/m\right)t}$ .

Explanation of Solution

Given information: The resistance force on an object of mass m moving with velocity v is thus −kv for some positive constant k.

Proof:

The required proved is obtained as,

  $\begin{array}{l}F=m\frac{dv}{dt}=-kv\\ \frac{dv}{v}=-\frac{k}{m}dt\\ v=v_0{e}^{-\frac{k}{m}t}\end{array}$

c.

To determine

To find the one is slow to half its starting velocity in the shortest time.

Answer to Problem 53E

A lighter mass will slow in time than a heavier mass.

Explanation of Solution

Given information: The resistance force on an object of mass m moving with velocity v is thus −kv for some positive constant k.

Calculation: The required solution is obtained as,

Substitute $v=\frac{1}{2}{v}_0$ into the equation from part (b) and solved for t,

  $\begin{array}{l}\frac{1}{2}{v}_0=v_0{e}^{-\frac{k}{m}t}\\ \frac{1}{2}=e^{-\frac{k}{m}t}\\ \ln \frac{1}{2}=-\frac{k}{m}t\\ \ln 1-\ln 2=-\frac{k}{m}t\\ -\ln 2=-\frac{k}{m}t\\ \frac{k}{m}t=\ln 2\\ t=\frac{m}{k}\ln 2\end{array}$

Therefore as m increase, t increase if k remains constant. The lighter mass with then slow down faster.