I'm stuck on this problem...what did i do wrong?
d(t)=(velocity) (t)+1/2 (acceleration) (t)^2
cycles 4 meters per second
rate of .4 meters per second squared
vertical distance from top to bottom of hill is 25 meters
How long until goes down the hill?
I have...
(1/5)t^2+4t-25
From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?