Expert: Scotto - 4/21/2009

Question
If an arrow is shot straight upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by s(t)=58t-0.83t^2

a)What is the velocity of the arrow in meters per second after 10 seconds?

b)How many seconds will it take for the arrow to return and hit the moon?

C)With what velocity in meters per second will the arrow hit the moon?



I don't understand.. don't you use the formula
(f(x)-f(a))
----------- ?? Becuase when I do, I can't simplify
  x-a      because i get a zero in the denom. Please help
           with formulas on how to do it.

Answer
The height of the arrow is s(t).
The velocity of the arrow is the derivative of s(t),
which I will call v(t), and it can be seen that since v(t) is the
derivative of s(t), v(t) = 58 - 1.66t.
Since there is an (f(x)-f(a))/(x-a) in here,

I'm not sure you've learned derivatives yet.
If the function is f(x), the derivative is f'(x).
So I will tell you.  If f(x)=x^n, f'(x)=nx^(n-1).
In other words, if f(x) = 4x, f'(x)=4.
If f(x)=5x², f'(x)=2*5x^1 = 10x.
If f(x)=7x^3, f'(x)=3*7x^(3-1) = 21x².

a) Putting t=10 in v(t) to find the velocity after 10 sec.

b) Since the moon is where s(t) = 0, set 58t-0.83t² equal to 0.
Note that a t factors out, giving t(58-0.83t)=0, so we get two answers.  The first is t=0, which is when it started.  The second is
58 = 0.83t, so divide 58 by 0.3 and you have the answer.

c) It will return with the reverse velocity of what is was shot at.
Note that on earth, we neglect air resistance.  On the moon, we don't even need to worry about air resistance since there is no air.
If you have read this far, I might as well say just take the
negative of how fast it was shot to get the return velocity.
A negative is how fast the arrow is rising.
If you take the negative of this (and get a positive),
the number gotten is fast it is coming down.