Question

Just a quick check -

http://prntscr.com/b9yfk8

Answer 1

5cos(2pi(t)/5)

Answer 2

Hm... let's see.
\(f(t)=5cos(2\pi(t/5))\)
f(0)=5cos(0)=5 (high point)
f(5)=5cos(2pi)=>period = 5 seconds
cos(2pi(t/5)) is a harmonic function.
f(2.5)=5 cos(pi)=-5
So the amplitude is 5-(-5)=10, not quite what was needed.
I am sure you can fix that!

Answer 3

@mathmate - so we're looking at

5cos(10pi t)?

Answer 4

For a function like
f(x)=Acos(2pi(t/5)), the range of f(x) is -A to +A because cos(x) can vary from -1 to +1.
In your question, it was clearly specified that the total distance (lowest to highest) is 5, that makes A=2.5.
The period remains at 5 seconds, so keep the 2pi(t/5).