How to Determine Maximum Acceleration of a Moving Object
Acceleration is the derivative of velocity. If a function gives the position of something as a function of time, you differentiate the position function to get the velocity function, and you differentiate the velocity function to get the acceleration function. Stated a different but equivalent way, the first derivative of position is velocity, and the second derivative of position is acceleration.
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, it’s 30 inches above the ground, and after 4 seconds, it’s at a height of 18 inches, as shown in the first graph in the figure.
Velocity, V(t), is the derivative of position (height, in this problem), and acceleration, A(t), is the derivative of velocity. Thus:
The graph of the acceleration function at the bottom of the figure is a simple line, A(t) = 6t – 12.
It’s easy to see that the acceleration of the yo-yo goes from a minimum of
at t = 0 seconds to a maximum of
at t = 4 seconds, and that the acceleration is zero at t = 2 when the yo-yo reaches its minimum velocity (and maximum speed). When the acceleration is negative — on the interval [0, 2) — that means that the velocity is decreasing. When the acceleration is positive — on the interval (2, 4] — the velocity is increasing.
An object is speeding up (what we call “acceleration” in everyday speech) whenever the velocity and the calculus acceleration are both positive or both negative. And an object is slowing down (“deceleration” in everyday speech) when the velocity and the calculus acceleration are of opposite signs.
Look at all three graphs in the figure again. From t = 0 to about t = 0.47, the velocity is positive and the acceleration is negative, so the yo-yo is slowing down while moving upward (till its velocity becomes zero and it reaches its maximum height). In plain English, the yo-yo is decelerating from 0 to about 0.47 seconds. The greatest deceleration occurs at t = 0 when the deceleration is
(the graph shows negative 12, but you can think of it as positive 12 because it is decelerating, get it?)
From about t = 0.47 to t = 2, both velocity and acceleration are negative, so the yo-yo is speeding up while moving downward. From t = 2 to about t = 3.53, velocity is negative and acceleration is positive, so the yo-yo is slowing down again as it continues downward (till it bottoms out at its lowest height). Finally, from about t = 3.53 to t = 4, both velocity and acceleration are positive, so the yo-yo is speeding up again. The yo-yo reaches its greatest acceleration of
at t = 4 seconds.