Question
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?
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?
Answers
Reiny
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
or
t = 100(100)/83 = appr 120.5 seconds
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
or
t = 100(100)/83 = appr 120.5 seconds