Find the terminal speed. Terminal speed means that, after a long time, the object’s speed becomes
constant. To find that terminal speed, do an equilibrium problem: Free body, set up forces equal down
forces, and left forces equal right forces. If something’s falling straight down with no other forces,
usually you’ll get bv = mg . Then solve for v . That’s the terminal speed.
- Sketch a graph of the speed of the object as a function of time. Perhaps the problem will say the
object was dropped from rest, or give an initial velocity. Well, you can plot that point at time t = 0.
Then you can find the terminal velocity using the method above—the terminal speed is the constant
velocity after a long time. Plot a horizontal line for the terminal velocity near the right-hand side of the
t -axis.
In between the initial velocity and the terminal velocity, just know that the velocity function will
look like an exponential function, changing rapidly at first, and changing less rapidly as time goes on.
Sketch a curve in between the point and the line you drew. Done. - Describe the motion in words, including what’s happening to the acceleration and/or the velocity.
Be sure to use a free-body diagram, and to separate the motion if necessary into parts when the object
is moving up, and moving down. When the speed is zero, the force of air resistance is zero. This
doesn’t mean no acceleration, of course. When the speed is not zero, use a free body to figure out the
amount and direction of the net force. Remind yourself of the basics of kinematics—the net force is the
direction of acceleration. If the acceleration is in the direction of motion, the object speeds up; if the
acceleration is opposite motion, the object slows down. - Solve a differential equation to find an expression for the velocity as a function of time. Again,
start with a free-body diagram, and write F (^) net = ma . Now, though, you’ll need to do some calculus:
acceleration . Perhaps your Newton’s second law equation might say something like mg – bv =
ma .
Okay, solve for a and write the calculus expression for . This type of equation is
called a differential equation, where a derivative of a function is proportional to the function itself.
Specifically, since the first derivative is involved, this is called a “first order” differential equation.
Your calculus class may well have taught you how to solve this equation by a technique known as
separation of variables: put all the v terms on one side, the t terms on the other, and integrate. If you
know how to do that, great; if not, it’s complex enough not in any way to be worth learning in order to
possibly—possibly —earn yourself one or two points. Everyone, though, should be able to recognize
and write the answer using the knowledge that the solution to a first-order differential equation will
involve an exponential function . You can use facts about the initial velocity and the terminal
velocity to write this function without an algorithmic solution.
Imagine that a ball was dropped from rest in the presence of air resistance F (^) air = bv . Writing the
second law gives you the equation shown above. What’s the solution? Well, the initial velocity is zero; the
terminal velocity can be found by setting acceleration to zero, meaning . The answer will
always be something like v (t ) = Ae – kt^ or v (t ) = A (1 − e − kt^ ). Start by figuring out which: does the
speed start large, and get small? If so, use the first expression. Or, does the speed start small, and get
larger? If so, use the second expression. In this case, the speed starts from zero and ends up faster. So we