Question

An object is moving in a straight line according to s(t) = -t^2+4t-3 where s(t) = distance the object goes in miles at. t=times in hours
a. How far did the object go in 2 hours?
b. How fast was the object in 3 hours?

Answer 1

(a) s(2)-s(0) = (-4+8-3)-(-3) = 4 mi
(b) v(t) = ds/dt = -2t+4
v(3) = -2*3+4 = -2 mi/hr

Answer 2

From problem 1a. ^
a. What was the acceleration at 3 hrs?
b. When was the object farthest away?

Answer 3

a(t) = -2, so constant
max s is when v=0