A model rocket is launched vertically upward. If it's position (height) y above the ground in feet as a function of time t in seconds is given by:
y = f(t) = 192t - 16t^2
The Attempt at a Solution
1. Find the instantaneous velocity and acceleration at any time t.
Now this is my first calculus course on my long academic journey, but velocity is the first derivative and acceleration is the second.
y' = 192-32t
That was easy, but it is the second derivative (acceleration) that is stumping me:
y'' = -32
How is that possible.. negative acceleration the entire time?
Thank you for your assistance.