Near the surface of the moon, the distance traveled by a falling object is given by the formula s(x) = (8t^2)/3, where t is measured in seconds in seconds, s(t) is in feet, and t = 0 corresponds to the instant that the object begins to fall.
Show that the average velocity over the interval [tsub1, tsub2] is given by
delta(s)/delta(t) = (8/3)(tsub2 + subt1)
Show that the average velocity over the interval [tsub1, tsub2] is given by
delta(s)/delta(t) = (8/3)(tsub2 + subt1)
-
s(x) = distance in feet
t = time in seconds
velocity in feet/sec = d[s(x)]/dt
s(x) = (8t^2)/3
ds(x)/d(t) = 16t/3
Average velocity over interval = (1/2)*(8/3)*[tsub2 - tsub1]
= (4/3)*[tsub2 - tsub1]
t = time in seconds
velocity in feet/sec = d[s(x)]/dt
s(x) = (8t^2)/3
ds(x)/d(t) = 16t/3
Average velocity over interval = (1/2)*(8/3)*[tsub2 - tsub1]
= (4/3)*[tsub2 - tsub1]