Question

An elevator starts from the first floor upwards. The height h in meters above the first floor after t seconds can be determined by the connection.

a) The elevator stops when the speed is zero. How high is the elevator then?
b) What is the acceleration of the elevator when it just started?

I've calculated A) to be 5.1425 m

I cannot figure out B) My calculation comes out to be -0.0008m/s^2 but I'm not sure that is right.

h '(t) = 1-2 * 0.18t
h '(t) = 1-0.36t

h '(t) = 1-0.36t = 0
1 = 0.36t
t = (1) / (0.36) = 2.78s

1-0.36 * 2.78 = -0.0008 m/s^2

Answer 1

if the velocity

v(t) = h '(t) = 1 - 0.36t
then the acceleration is
a(t) = v'(t) = -0.36

hard to say, since you don't say what "the connection" is.