The height of an object t seconds after it is dropped from a height of 300 meters is s(t)=-4.9t^2 +300. Find the average velocity of the object during the first 3 seconds? (b) Use the Mean value Theorem to verify that at some time during the first 3 seconds of the fall the instantaneous velocity equals the average velocity. Find that time.
Average Velocity:
[ f(3) - f(0) ] / ( 3 - 0 )
Calculate f(3):
f(3) = -4.9(3^2) + 300
f(3) = -4.9(9) + 300
f(3) = -44.1 + 300
f(3) = 255.9
Calculate f(0):
f(0) = -4.9(0^2) + 300
f(0) = 0 + 300
f(0) = 300
So we have average velocity:
Average velocity = (255.9 - 300)/(3 - 0)
Average velocity = -44.1/3
Average velocity = -14.7
Velocity is the first derivative of position
s(t)=-4.9t^2 +300
s'(t) = -9.8t
So we set velocity equal to average velocity:
-9.8t = -14.7
Divide each side by -9.8 to solve for t, we get t = 1.5
Average Velocity:
[ f(3) - f(0) ] / ( 3 - 0 )
Calculate f(3):
f(3) = -4.9(3^2) + 300
f(3) = -4.9(9) + 300
f(3) = -44.1 + 300
f(3) = 255.9
Calculate f(0):
f(0) = -4.9(0^2) + 300
f(0) = 0 + 300
f(0) = 300
So we have average velocity:
Average velocity = (255.9 - 300)/(3 - 0)
Average velocity = -44.1/3
Average velocity = -14.7
Velocity is the first derivative of position
s(t)=-4.9t^2 +300
s'(t) = -9.8t
So we set velocity equal to average velocity:
-9.8t = -14.7
Divide each side by -9.8 to solve for t, we get t = 1.5