Solving a Differential Equation for Bungee Jumping Motion

[OrisAble]

Well as above I've completely hit a blank for some uni work I've got.
If anyone could give me an idea of what i actually need to do here that would be great. I don't need someone do this for me. Just clear up what it is they want me to do.

"...Bungee jumping situation

Choose x= x(t) (positive donwards) to represent the location of the jumper's feet below the launching platform. Assume the bungee cord is of length L = 50m. Hooke's Law is to be used to represent the restoring force when the cable is stretched beyond length L, that is

Restoring force =
{0 x<=L
{k(x-L) x > L

Assume further, that their is an air resistance term opposing the motion of magnitude

cx(with the dot on top)|x(with the dot on top)|

Use Newton's equation of motion to determine the DE that describes the motion at times t>=0."

Thats what i don't understand. What does that actually want?[/quote]
Newton's equation is "force= mass times acceleration" or, for constant mass,
$$m\frac{d^2x}{dt^2}= F$$
Here you are told that F is the sum of two separate forces: the restoring force which is 0 for x< L and -k(x-L) for x>L (the "-" is because the force is back toward x= 0) and the air resistance force which is -c dx/dt (again negative because the air resistance is always opposite to the direction of motion). Put those together.

Would be great if anyone could give me some direction..btw, ill be replying with a different account as i had forgotten my password and can't check hotmail accounts at uni ;).

 

[OrisAble]

well i just found out my account was deleted. Just realized i signed up with the same name as i use to have. maybe some of you remember me...

 

[HallsofIvy]

Newton's law is "force equals mass times acceleration" or, for constant mass, m d²x/dt²= F. Here you are told that the force is the sum of two forces: the restoring force, 0 if x< L, -k(x-L) (negative since the force is back toward x=0) and the air resistance force -c dx/dt (again negative since resistance is always opposite the direction of motion). Put those together.