Velocity Dependent Force Problems
You will encounter problems where one of the forces acting on an object is a constant times the velocity
of the object, and you are asked to find the velocity of the object. The strategy to use is to “make a dot”
and then apply Newton’s second law (N2L). To solve the problem, however, you will need to recast N2L
in the form of a first order differential equation and solve it either by guessing the solution, or by using
integration. Please note from the following worked example that you must show all of your steps to
earn all of the points. Go through the following example step by step making notes in the margins and
then do the problems that follow at the end.
Example from a past AP Exam:
A box of mass m initially at rest is acted upon by a constant applied force of magnitude FA, as shown in
the figure below. The friction between the box and the horizontal surface can be assumed to be
negligible, but the box is subject to a drag force of magnitude kv where v is the speed of the box and k is
a positive constant.
Express all your answers in terms of the given quantities and fundamental constants, as appropriate.
(a) The dot (right) represents the box. Draw and label the forces (not
components) that act on the box.





For correctly showing and labeling the applied force directed to the
right
1 point
For correctly showing and labeling the downward gravitational
force
1 point
For correctly showing and labeling the upward normal force 1 point
For correctly showing and labeling the drag force directed to the
left
1 point
One earned point was deducted for having any extraneous vectors
(b) Write, but do not solve, a differential equation that could be used to determine the speed v of the
box as a function of time t. If you need to draw anything other than what you have shown in part (a)
to assist in your solution, use the space below. Do NOT add anything to the figure in part (a).
For the correct substitution into Newton’s second law
1 point
∑ 𝐹 = 𝑚𝑎
𝐹𝐴 − 𝑘𝑣 = 𝑚𝑎
For a correct differential equation
1 point
Simply substitute the derivative of the velocity for the acceleration.
𝑑𝑣
𝐹𝐴 − 𝑘𝑣 = 𝑚
𝑑𝑡
(c) Determine the magnitude of the terminal velocity of the box.
Terminal velocity occurs when the velocity is no longer changing. This means that the change in
velocity with time is zero, so:
𝑑𝑣
𝐹𝐴 − 𝑘𝑣 = 𝑚 𝑑𝑡
…. Simply set
𝑑𝑣
𝑑𝑡
= 0, so:
𝐹𝐴 − 𝑘𝑣 = 0
For the correct expression for the terminal velocity
𝑣𝑇 =
1 point
𝐹𝐴
𝑘
(d) Use the differential equation from part (b) to derive the equation for the speed v of the box as a
function of time t. Assume that v = 0 at time t = 0.
Note the use of the word “derive”. This means we can’t just guess the solution, but if you did you
would get partial credit. The proper way to derive the equation is to use the differential equation
from part (b) and integrate using separation of variables.
𝑑𝑣
𝐹𝐴 − 𝑘𝑣 = 𝑚
𝑑𝑡
For demonstrating separation of variables
1 point
1
1
𝑑𝑡 =
𝑑𝑣
𝑚
𝐹𝐴 − 𝑘𝑣
For demonstrating that the equation must be integrated
∫
1 point
1
1
𝑑𝑡 = ∫
𝑑𝑣
𝑚
𝐹𝐴 − 𝑘𝑣
For demonstrating substitution using initial and final values (or evaluating the
constant of integration using the boundary conditions)
𝑡
∫
0
1 point
𝑣(𝑡)
1
1
𝑑𝑡 = ∫
𝑑𝑣
𝑚
𝐹𝐴 − 𝑘𝑣
0
1
Note: set 𝑢 = 𝐹𝐴 − 𝑘𝑣 … then 𝑑𝑢 = −𝑘𝑑𝑣 … 𝑑𝑣 = − 𝑘 𝑑𝑢
𝑣(𝑡)
∫
0
1
1 𝑢(𝑡) 1
𝑑𝑣 = − ∫
𝑑𝑢 … 𝑡ℎ𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑎𝑠 𝑦𝑜𝑢 𝑛𝑜𝑟𝑚𝑎𝑙𝑙𝑦 𝑤𝑜𝑢𝑙𝑑
𝐹𝐴 − 𝑘𝑣
𝑘 𝑢(0) 𝑢
… substitute back into −
1
ln(𝑢)
𝑘
…
𝑎𝑛𝑑 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑛𝑑 𝑓𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
(… or you may use the approach where you solve for the constant of integration)
𝑡 𝑡
1
𝐹𝐴 − 𝑘𝑣 𝑣
)]
] = − [𝑙𝑛 (
𝑚 0
𝑘
1
𝑜
For attempting to solve for v(t)
−
1 point
𝑘𝑡
𝐹𝐴 − 𝑘𝑣
= 𝑙𝑛 (
)
𝑚
𝐹𝐴
𝑘𝑡
𝑒−𝑚 =
𝐹𝐴 − 𝑘𝑣
𝑘𝑣
=1−
𝐹𝐴
𝐹𝐴
𝑘𝑡
𝑘𝑣
= 1 − 𝑒 −𝑚
𝐹𝐴
For a correct answer
𝑣=
1 point
𝑘𝑡
𝐹𝐴
(1 − 𝑒 − 𝑚 )
𝑘
(e) On the axes below, sketch a graph of the speed u of the box as a function of time t. Explicitly label
any intercepts, asymptotes, maxima, or minima with numerical values or algebraic expressions, as
appropriate.
For a graph that begins at the origin, with a non-negative slope everywhere,
and is concave downward
1 point
For a graph with a horizontal asymptote
1 point
For the correct label of the expression for the asymptote or maximum
on the vertical axis
1 point
Your Turn
Solve the following similar problems:
Notice that in this problem there is no applied force, and the initial conditions are different so the object
is slowing down rather than speeding up:
1. A generic body of mass m is moving along the x-axis with velocity v, and is then slowed by a
force in the opposite direction to the motion, F = -kv, where k is a constant. At time t = 0, the
object has velocity vo at position x = 0.
a. What is the initial acceleration (magnitude and direction) produced by the resistance
force?
b. Derive an equation for the object's velocity as a function of time t, and sketch this
function on the axes below. Let a velocity directed to the right be considered positive.
Hint: This problem is easier since you can directly integrate the expression without
substituting variables.
c. Derive an equation for the distance the object travels as a function of time t and sketch this function
on the axes below. Hint: Now you need to show velocity as the derivative of the position, and
integrate the expression for the velocity to obtain the expression for position.
d. Determine the distance the object travels from t = 0 to t = .
2. An object of mass m near the Earth’s surface falls from rest in the Earth's gravitational field. Acting
on the object is a resistive force of magnitude kmv, where k is a constant and v is the speed of the
object.
a. On the diagram below, draw and identify all of the forces acting on the object as it falls.
b. Write the differential equation that represents Newton's second law for this situation.
c. Determine the terminal speed vT of the object.
d. Integrate the differential equation once to obtain an expression for the speed v as a function of
time t. Use the condition that v = 0 when t = 0.
e. On the axes provided below, draw a graph of the speed v as a function of time t.