Posted by Linda on Saturday, June 8, 2013 at 10:12pm.
If an arrow is shot straight up from the surface of the moon with an initial velocity of 100 ft/s, its height in feet after t second is given by s(t)=(100t)−(83/100)(t^2).
Use the limit definition of the derivative to find the answers to the following questions.
Find the velocity of the arrow when t=2.
Find the velocity of the arrow when t=a.
When will the arrow return to the moon's surface?
What will the velocity of the arrow be when it hits the surface?
- Calculus - Reiny, Saturday, June 8, 2013 at 10:53pm
s(2) = 100(2) - (83/100)(4) = 4917/25
s(2+h) = 100(2+h) - (83/100)(2+h)^2
= 200+100h - 332/100 - (332/100)h - (83/100)h^2
velocity = Limit ( s(2+h) - s(2) )/(2+h - 2) as h -->0
= Lim (200+100h - 332/100 - (332/100)h - (83/100)h^2 - 4917/25)/h
= Lim ( (2417/25)h - (83/100)h^2 )/h
= lim 2417/25 - (83/100)h , as h ---> 0
= 2417/25 ft/sec
carefully repeat the above steps using a instead of 2
you should get
v(a) = 100 - (83/50) a
the arrow will return to the surface when s(t) = 0
100t - (83/100)t^2 = 0
t (100 - (83/100)t ) = 0
t= 0 ----> the start of the shot
t = 100(100)/83 = appr 120.5 seconds
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