Question

Calculus with bungee cords help! 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.

Answer 1

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.

Answer 2

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)?

Answer 3

Attempt at problem 1 (There is more to this I just want to confirm my answers thus far):
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: Problem 2: H(0)= -415 ft
V(0)= -113 ft/s
A(0)= -32 ft/s2

Answer 4

This sounds like Damped Harmonic Motion, which has a 2nd order differential equation. Are you studying diff eq ?

Answer 5

@phi No I'm in AP Calculus BC and even though I'm taking physics we're not really supposed to know any. This is supposed to be done using calculus. In a little bit, if you're still helping me, I need someone to confirm/help me with this using Euler's method.

Answer 6

@phi I guess we kinda are studying diff eq. but not so much.

Answer 7

@phi Screw it, this will make it easier. https://drive.google.com/file/d/0B8z1pN15n_d8d2VibGxaNXFzSHc/view?usp=sharing

Answer 8

The acceleration of the jumper is not just 32 ft/sec^2
using the idea F= m a of the jumper, and the equation that the force is
-mg (force is down, so negative)
- v (assuming velocity is negative, so -v will be positive i.e. up. This is the force due to wind resistance... notice this reduces the force due to gravity)
and finally
- 0.39* s where s is the amount of "stretch" in the cord.

when the jumper is at distance L below the bridge, s is exactly 0 (no stretch on the cord)

thus the formula for the total force is
F= - m g - v
or , replacing F with m a:
ma = -m g - v
divide by the mass:
a = -g -v/m

putting in g= 32 ft/sec^2 and v= -113, and m = 5 we get
a= -32 + 113/5
a = -9.4 ft/sec^2
at time = 0 (just when the jumper is L ft below the bridge)

Answer 9

for Problem 2
we have a small time step dt = 0.1 second
During that short time period, we assume the velocity is constant, so we can say the distance fallen will be Velocity * dt
H(j+1) = H(j) + dt*V(j)
similarly the velocity at the next time step will be
V(j+1) = V(j) + dt*A(t) (neg means down in this case)

Acceleration is a bit more complicated.

Answer 10

I will let you post what you think the formula is for
A(j+1) = ???

Meanwhile, here is a graph of the distance fallen as a function of time:

Answer 11

@phi Ok yeah that makes sense about the acceleration... I think I was thinking the acceleration they gave us took into account air resistance.

The A(j) equation would be A(j+1)=A(j)+.1*A'(j). A(j) can be re-written as \[A(j)=\frac{ -32m+0.39(-L-H(j))-V(j) }{ m }\]
Thus A'(j) will be (L is a constant) \[A'(j)= -0.078H'(j)-0.2V'(j)\]
Which simplifies to
\[A'(j)=-0.078V(j)-0.2A(j)\]
So we would use A(j) for that and that for A'(j) for the original equation of \[A(j+1)=A(j)+0.1A'(j)\]

Answer 12

yes, that looks ok

Answer 13

@phi ok then I shouldn't need help with problem 4. Give me a little bit and I'll try to do problem 6 and on. Before I couldn't turn off the acceleration graph but I'll deal with

Answer 14

@phi For some reason I can't turn off the third graph using Fnoff

Answer 15

I never use a graphing calculator. They do come with a (very long) manual, which you can download from the web. If you have excel, that is easier to control... but like anything it takes time to learn.

Answer 16

@phi What do you use? Is there a good online graphing calculator for these?

Answer 17

I use matlab (or octave which is a free download)
but it takes a bit of time to learn.

Answer 18

@phi Do I need those to do problem 6 and on? Also I think A(0) is -32 because they even said it was in the part where they are putting everything into the calculator.

Answer 19

you will need a calculator at least.
Yes, they screwed up A(0)= -32 is not correct. (unless we ignore the wind resistance, but they say we should include it). if you are falling at -113 ft/sec the wind is slowing you down and we are not accelerating at -32 ft/sec-sec

Answer 20

@phi Yeah they did screw up, before the graph would not show entirely but putting in the acceleration you calculated it does, this is why you are at MIT lmao. Thank you so much man

Answer 21

@phi Does this look right for V(n)?

Answer 22

@phi Does this look right for A(n)?

Answer 23

yes, both V and A plots have the correct shape.
Here are matlab versions.

Answer 24

@phi Ik I'm asking for a lot but can you just visit this: https://docs.google.com/presentation/d/18hf75l-ngUP1TpnFiGTe95UX7FsCsKnkDZi0ocOX6oE/edit#slide=id.g1320c55de2_0_32

And look through the last problems and make sure I have them correct?

Thank you so much for all your help. <3

Answer 25

I assume you know that the slides with just answers is not "stand-alone" and is not understandable without knowing the details of the question. In other words, if you read this a year from now, you will not know what you did.

A "nice" presentation would give the problem, define the coordinate system,
a short review of the physics, etc...

In problem 2, you have a typo: your equation should read
ma=-32m-v
(rather than ma=-32m+v)
because v will be negative (going down) and you want the resistance to act "upward" i.e. be positive.

in problem 7, I got about 826 ft for the steady-state. But your answer is probably close enough.