Final answer:
To find the velocity and speed of the potato at specific times after being launched, one must derive the height function and evaluate at those times. The velocity is positive when height is increasing and negative when it is decreasing. By finding when the velocity is zero, we can determine the intervals of increase and decrease.
Step-by-step explanation:
The question involves finding the velocity and the speed of the potato at certain times after it has been launched from a potato cannon, as well as determining the intervals during which the potato's height is increasing and decreasing.
The velocity of the potato at any time t can be found by taking the derivative of the height function s(t). The height function given is s(t) = -16t2 + 105t + 90. The derivative of this function, s'(t) = -32t + 105, represents the velocity of the potato at time t.
To find the velocity at t = 0.5 seconds, we substitute t with 0.5: s'(0.5) = -32(0.5) + 105 = -16 + 105 = 89 ft/s. Similarly, for t = 5.5 seconds: s'(5.5) = -32(5.5) + 105 = -176 + 105 = -71 ft/s. The speed is the absolute value of the velocity, so the speed at t = 0.5 seconds is 89 ft/s and at t = 5.5 seconds is 71 ft/s.
The potato's height is increasing and decreasing based on the sign of the velocity. If the velocity is positive, the height is increasing; if the velocity is negative, the height is decreasing. By finding the time when the velocity is zero, we can determine these intervals. The velocity is zero when s'(t) = 0, which occurs at t = 3.28 seconds. Therefore, the height is increasing on the open interval (0, 3.28) seconds and decreasing on the open interval (3.28, t where s(t) hits the ground).