How to Determine Maximum and Minimum Speeds of Moving Objects - dummies

How to Determine Maximum and Minimum Speeds of Moving Objects

By Mark Ryan

One of the most practical uses of differentiation is finding the maximum or minimum values of a real-world function, for example, the maximum and minimum speeds of a moving object.

You can think of velocity as the more technical version of speed.

Here’s an example. A yo-yo moves straight up and down. Its height above the ground, as a function of time, is given by the function H(t) = t3 − 6t2 + 5t + 30, where t is in seconds and H(t) is in inches. At t = 0, the yo-yo is 30 inches above the ground, and after 4 seconds, it’s at a height of 18 inches, as shown in this figure.

The yo-yo's height, from 0 to 4 seconds.
The yo-yo’s height, from 0 to 4 seconds.

To determine the total distance the yo-yo travels, you need to add up the distances traveled on each leg of the yo-yo’s trip: the up leg, the down leg, and the second up leg.

First, the yo-yo goes up from a height of 30 inches to about 31.1 inches (where the first turn-around point is). That’s a distance of about 1.1 inches. Next, it goes down from about 31.1 to about 16.9 (the height of the second turn-around point). That’s a distance of 31.1 minus 16.9, or about 14.2 inches. Finally, the yo-yo goes up again from about 16.9 inches to its final height of 18 inches. That’s another 1.1 inches. Add these three distances to obtain the total distance traveled:


Note: Compare this answer to the total displacement of  12, which you get from subtracting the final height of the yo-yo, 18 inches, from its initial height of 30 inches. The displacement is negative because the net movement is downward. And the positive amount of the displacement (namely 12) is less than the distance traveled of 16.4 because with displacement the up legs of the yo-yo’s trip cancel out part of the down leg distance. Check out the math:


The yo-yo’s average speed is given by the total distance traveled divided by the elapsed time. Thus,


Say you determine that the yo-yo’s maximum velocity is 5 inches per second, and its minimum velocity is –7 inches per second. A velocity of –7 is a speed of 7, so that’s the yo-yo’s maximum speed. Its minimum speed of zero occurs at the two turnaround points.

A good way to analyze maximum and minimum speed is to consider the speed function and its graph. (Or, if you’re a glutton for punishment, check out the following mumbo jumbo.) Speed equals the absolute value of velocity.

Velocity, V(t), is the derivative of position (height, in this problem). Thus:


So, for the yo-yo problem, the speed function,


Check out the graph of S(t) in the following figure.

Looking at this graph, it’s easy to see that the yo-yo’s maximum speed occurs at t = 2


and that the minimum speed is zero at the two x-intercepts.

Minimum and maximum speed: For a continuous velocity function, the minimum speed is zero whenever the maximum and minimum velocities are of opposite signs or when one of them is zero. When the maximum and minimum velocities are both positive or both negative, then the minimum speed is the lesser of the absolute values of the maximum and minimum velocities. In all cases, the maximum speed is the greater of the absolute values of the maximum and minimum velocities. Is that a mouthful or what?