At lift off space shuttle discovery had the constant acceleration, a, of 16.4 ft/sec2. Initial velocity, v 0, due to rotation of earth os 1341 ft/sec. Utilize function of
d(t) = v0t + 1/2 at2 to find distance from earth for every interval, t, after liftoff
a) 30 sec
b) 1 min
c) 2 min
d) if time space shuttle is in flight doubles, does distance from earth double. Why?