Question

Let s(t) = 512 - 22t be the position function of a particle moving along a coordinate line, where s is in feet and t is, in seconds. Find the maximum speed of the particle during the time interval 1 < t < 3.

When, during the time interval 1

Answer 1

The maximum speed of the particle with position function s(t) = 512 - 22t is 22 ft/s, since its velocity function v(t) = -22 ft/s indicates a constant speed throughout the motion during the time interval 1 < t < 3.

Step-by-step explanation:

To find the maximum speed of the particle when its position function is given by s(t) = 512 - 22t, we need to find the first derivative of s(t), which represents the velocity function v(t). The maximum speed occurs when the magnitude of velocity is greatest. The velocity function v(t) is obtained by differentiating the position function with respect to time:

v(t) = -22 ft/s

This velocity function indicates a constant velocity, which means the speed is constant throughout the motion, and the maximum speed is thus equal to 22 ft/s for the given time interval 1 < t < 3.

To find the maximum speed of the particle during the time interval 1 < t < 3, we need to first find the velocity function v(t). The velocity function is the derivative of the position function, so we differentiate s(t) = 512 - 22t with respect to t. Differentiating, we get v(t) = -22 m/s. Since we want to find the maximum speed, we take the absolute value of v(t), giving us |v(t)| = 22 m/s.