The previous section introduced a complication into the study of an object in free fall by taking air resistance into account. In this section, we will find the falling speed of the object as a function of time using a differential equation. This provides an analytic solution to the problem of motion involving non-constant acceleration due to air resistance, and is a contrast to the numerical analysis (using a spreadsheet) that you saw in the previous section. The solution provides an indication of the mathematical sophistication needed to consider motion with non-constant acceleration.

The first equation in Equation 1 is from the last section, and it can be rearranged into the form Δv/Δt = g − kv_p². The equation below it is an equivalent differential equation, which we will solve. The solution will be a function expressing the speed of a falling object in terms of time.

In this section, as in the last, we take the downward direction as positive (instead of the usual negative) for the sake of simplicity, so the speed and acceleration of the falling object will both be positive. A little simplicity will not hurt in this section!

The expression on the left side of the differential equation is the time derivative of speed, which is acceleration. In the absence of air resistance, the acceleration would be g, but with air resistance, it has a smaller magnitude. This is reflected in the terms on the right side of the equation.

Variables

Strategy

1. The differential equation in Equation 1 is known as a separable differential equation because the two variables, v and t, can be moved to opposite sides of the equation. Rearrange the differential equation in this form.
2. Integrate both sides of the differential equation, ending up with an arbitrary constant of integration. Solve the resulting equation for the speed v as a function of time.
3. Use the information stated below about the initial velocity of the object to establish a numerical value for the constant. This results in an equation that gives the speed of the object as a function of time.

Physics principles and equations

The acceleration of the falling object equals the acceleration of gravity minus the drag of air resistance.

The object starts falling from rest, so its initial velocity is zero.

v(0) = 0

Mathematics facts

Exponentiation and taking logarithms are inverse operations.

e ^ln(x) = x

The rule for multiplying powers of a common base.

e ^a e ^b = e ^a + b

Step 5 of the derivation requires an integral that can be found in a table of integrals.

Using elementary algebra, the first of the following two equations can be solved for x, as shown in the second.

Step-by-step derivation

The equation above can be compared to the results from the spreadsheet. The units of k have to be inverse meters (m⁻¹), since k multiplied by v² (which has units of m²/s²) must have the units of acceleration (m/s²). In order to compare the analytic solution to the solution given by the spreadsheet, the value of the constant k can be taken to match its default value in the spreadsheet in the prior section, 0.200 m⁻¹.

Using this value for k, the speed after, say, 3.62 seconds is 7.00 m/s, calculated using either the formula above, or using the spreadsheet. In the spreadsheet this was the terminal velocity of the falling object. Letting t → ∞ in the analytic formula shows that the terminal velocity is the square root of g/k: 7.00 m/s.

In this section, the differential equation is (relatively) simple, so that an analytic solution can be found. In more complicated models of motion, analytic solutions are much more difficult, or impossible, and numerical simulations are then a physicist’s best friend.