# Calculus with bungees

## Homework Statement

So this involves a little physics but it is to all be done using calculus (They give us everything we need). Everything is in terms of feet so acceleration due to gravity is 32 ft/(s^2) they tell us the mass of the jumper is 5 slugs. They then tell us that the Force due to the cord of bungee is Fc=0.39s where s is the length of the stretched part of the cord (basically just like the force of a spring). -L is the equilibrium point, basically where the bungee cord is at it's full length below the bridge but unstretched. Finally there is the force of air resistance but they just tell us it is proportional to the speed but in the opposite direction so Far= -v.

Problem 1: Let h=0 be the level of the bridge and fix our coordinate system so that the bottom of the gorge corresponds to h= -1053. At the instant the jumper reaches L feet below the bridge his height is h= -L. Let h<0 be the position of the jumper. As long as h<= -L, the total force F of the three forces on the jumper is F= -32m+0.39(-L-h)-v Find the acceleration where m=5 slugs.

Also we had three bungee cords to choose from, I chose the 415 ft cord where it's speed is 113 ft/s L feet below the bridge.

They then tell us that at t=0 is the time at which the jumper is L feet below the bridge.

Problem 2: For t>=0 let H(t) be the position of the jumper at time t, V(t) his velocity at time t, and A(t) be the acceleration at time t. What is H(0), V(0), and A(0)?
2. The attempt at a solution
Problem 1: I'm pretty sure I have this one I just want to start here and confirm my answers. The acceleration is denoted as a=(-32m+0.39(-L-h)-v)/m we can then plug in L and m (h and v are constantly changing) so a= (-321.85-0.39h-v)/5.

Problem 2: H(0)= -415 ft
V(0)= -113 ft/s
A(0)= -32 ft/s2

Please confirm these answers and then we can move on to the real calculus. Thanks!

Simon Bridge
Homework Helper
I'll annotate the problem statement ... this is how I'm reading it.

## Homework Statement

So this involves a little physics but it is to all be done using calculus (They give us everything we need). Everything is in terms of feet so acceleration due to gravity is 32 ft/(s^2) they tell us the mass of the jumper is 5 slugs. They then tell us that the Force due to the cord of bungee is Fc=0.39s where s is the length of the stretched part of the cord (basically just like the force of a spring). -L is the equilibrium point, basically where the bungee cord is at it's full length below the bridge but unstretched. Finally there is the force of air resistance but they just tell us it is proportional to the speed but in the opposite direction so Far= -v.
The bungee is stretched at the equilibrium point... otherwise how is it an "equilibrium"?
You have F=ks where s is the extension and k=0.39 slugs/s2 (guessing units) The equilibrium point is where this force matches the weight of the jumper.
Your reasoning is that the jumper loses gravitational potential energy, exchanging for kinetic energy (and work to overcome air resistance) until the unstretched length is reached (is terminal velocity reached first?) ... then kinetic and potential energy is exchanged for extension in the chord until there is none left.

Problem 1: Let h=0 be the level of the bridge and fix our coordinate system so that the bottom of the gorge corresponds to h= -1053. At the instant the jumper reaches L feet below the bridge his height is h= -L. Let h<0 be the position of the jumper. As long as h<= -L, the total force F of the three forces on the jumper is F= -32m+0.39(-L-h)-v Find the acceleration where m=5 slugs.
... oh I hate doing things in negative numbers: I always lose a minus sign somewhere.

Also we had three bungee cords to choose from, I chose the 415 ft cord where it's speed is 113 ft/s L feet below the bridge.
So v is specified for a particular position? That would be useful.

They then tell us that at t=0 is the time at which the jumper is L feet below the bridge.
... so if the position of the jumper is h, then h(0)=-L and ##\dot h(0)=-113##ft/s ... are they expecting you to solve a second order differential equation.

Problem 2: For t>=0 let H(t) be the position of the jumper at time t, V(t) his velocity at time t, and A(t) be the acceleration at time t. What is H(0), V(0), and A(0)?
... and changing variables again!!
Oh well... they tell you that ##H(0)=-L##, ##V(0)=\dot H(0)## which you chose when you picked the chord, and ##A(0) = ## from problem 1.

2. The attempt at a solution
Problem 1: I'm pretty sure I have this one I just want to start here and confirm my answers. The acceleration is denoted as a=(-32m+0.39(-L-h)-v)/m we can then plug in L and m (h and v are constantly changing) so a= (-321.85-0.39h-v)/5.
You are given ##v=\dot h## here.

For t<0 ##\dot H - mg = m\ddot H## ... second order DE.
Fortunately you don't have to solve it (yet) ... ##A(0)=\frac{1}{m}(V(0)-mg)##

Problem 2: H(0)= -415 ft
V(0)= -113 ft/s
A(0)= -32 ft/s2

Please confirm these answers and then we can move on to the real calculus. Thanks!
fwiw ... that's how I'm reading it.

I'll annotate the problem statement ... this is how I'm reading it.

The bungee is stretched at the equilibrium point... otherwise how is it an "equilibrium"?
You have F=ks where s is the extension and k=0.39 slugs/s2 (guessing units) The equilibrium point is where this force matches the weight of the jumper.
Your reasoning is that the jumper loses gravitational potential energy, exchanging for kinetic energy (and work to overcome air resistance) until the unstretched length is reached (is terminal velocity reached first?) ... then kinetic and potential energy is exchanged for extension in the chord until there is none left.

... oh I hate doing things in negative numbers: I always lose a minus sign somewhere.

So v is specified for a particular position? That would be useful.

... so if the position of the jumper is h, then h(0)=-L and ##\dot h(0)=-113##ft/s ... are they expecting you to solve a second order differential equation.

... and changing variables again!!
Oh well... they tell you that ##H(0)=-L##, ##V(0)=\dot H(0)## which you chose when you picked the chord, and ##A(0) = ## from problem 1.

You are given ##v=\dot h## here.

For t<0 ##\dot H - mg = m\ddot H## ... second order DE.
Fortunately you don't have to solve it (yet) ... ##A(0)=\frac{1}{m}(V(0)-mg)##

fwiw ... that's how I'm reading it.
Hey Simon, thanks for the reply.
We managed to figure everything out except we have a last part to the overall project.
Essentially we need to find the velocity of a 4 slug mass as it is dropped from the bridge and falls 415 feet below it. This confuses me because the force acting on the mass is 32m-v where the velocity at each point is literally the force of the air resistance therefor the acceleration we need to find is a=(32m-v)/m and we need to use this to solve for the final velocity 415 feet below it probably using kinematics. Can you help?