calculus
posted by Anonymous on .
a key is dropped from a height of 30 m , If its distance from the ground at t seconds after the release is given by the position function d (t) = 305tÂ², find
a.its height velocity and acceleration after 1 second.
b. the time it reached the ground.
c. its position after 1 second

If d(t) = 30  5t^2 m, then the velocity is
v(t) = 10t m/sec , and the acceleration is
a(t) = 10 m/sec^2
(each is the derivative of the previous function)
a) when t = 1
distance = 30  5(1^2) = 25 m
velocity = 10(1) = 10 m/s
acceleration = 10 m/s^2
b) when it reaches the ground, d(t) = 0, so
0 = 30  5t^2
t^2 = 6
t = √6 or appr. 2.4 seconds
c) 25 m above the ground , already answered in a)