Recent content by Sheen91

Homework Statement [PLAIN]http://img697.imageshack.us/img697/6702/bungee.png Derive the equations of the motion for the jumper. Consider only the vertical motion of the jumper. (g = 9.81 m/s^2) Homework Equations Straing = (Stated Above) Strain Rate = v / lo = v/25 F = -kx...

There are two answer rads/s and degs/s, both are not the same as my answer. I definitely did something wrong.

I manage to get the question. Thanks :D

Homework Statement http://img707.imageshack.us/img707/2066/question3.jpg Homework Equations v = r' \hat{r} + r*θ' (^θ) The Attempt at a Solution I was sure I knew how to do this question but made a mistake somewhere, because my answer is not right. So here is my working out. l r' l =...

Homework Statement [PLAIN]http://img231.imageshack.us/img231/636/question2m.jpg Homework Equations v = \dot{r} \hat{r} + r\dot{\vartheta} \hat{\vartheta} The Attempt at a Solution \dot{r} = ? \vartheta = 80° v = 55mm/s So I guess I just use the formula above. v = \dot{r} \hat{r} +...

I think I got it: Product rule 2y dy/dt = (24x² + 8) dx/dt (2 dy/dt) . dy/dt + d²y/dt² = (48x . dx/dt) + ( (24x² + 8) . d²x/dt²) so that means: d²y/dt² = (48x . dx/dt) + ( (24x² + 8) . d²x/dt²) - (2 dy/dt) . dy/dt d²y/dt² = -4791.58 That seems very wrong. Also after you get d²y/dt² and...

I am unsure on how to find the "Normal component of the Acceleration" Vela can you let me know what to do step by step, rather than just one step at a time. I would really apreciate that. I also got the numerical values of y and y', and am once again stuck. Also how do I use the product rule...

y = sqrt (8x³ + 8x) = sqrt (8.4³ + 8.4) = sqrt (544) = 23.324 y' = (24x² + 8) x' / 2y = 58.82 now what?

Sorry I don't understand how, do I use the product rule in that?? Thanks for all the help so far vela, I really appreciate it EDIT: uhmm ohh wait dy/dx = (dy/dt) / (dx/dt) 2y dy/dt = (24x² + 8) dx/dt dy/dx = (dy/dt) / (dx/dt) = (24x² + 8) / 2y d²y/dx² = ( (48x . 2y) - 2(24x² + 8)dy/dx ) /...

y² = 8x³ + 8x 2y dy/dt = (24x² + 8) dx/dt Velocity = sqrt ( (dx/dt)² + (dy/dt)² ) Acceleration = sqrt ( (d²x/dt²)² + (d²y/dt²)² ) = ? :S (not sure) 2y dy/dt = (24x² + 8) dx/dt 2 d²y/dt² = 48x d²x/dt² d²y/dt² = 24x d²x/dt² Therefore: Acceleration = sqrt ( (d²x/dt²)² + (d²y/dt²)² ) =...

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8y² + 4y = -4x + 6 sub in : "x = 0.75 - y" 8y² + 4y = -4(0.75 - y) + 6 8y² + 4y = -3 + 4y + 6 8y² = 3 y = sqrt(3/8) - because y > 0 If I sub that value into dx/dt and dy/dt dy/dt = -(1 / y) dy/dt = -(1 / sqrt(3/8)) = -1.633 dx/dt = 4 + (1 / y) dx/dt = 4 + (1 / sqrt(3/8)) = 5.633...

fixed I am still a little stuck at the same point

Ok so 8y² + 4y = -4x + 6 16y dy/dt + 4 dy/dt = -4 dx/dt (16y + 4) dy/dt = -4 . dx/dt sub in dx/dt = 4 - dy/dt (16y + 4) dy/dt = -4 (4 - dy/dt) (16y + 4) dy/dt = -16 + 4 dy/dt (16y) dy/dt = -16 (y) dy/dt = -1 dy/dt = -(1 / y) And dx/dt = 4 - dy/dt dx/dt = 4 + (1 / y) Now I we have...

x = 4t - y dx/dt = 4 (because then y would be a contant.) You want me to sub in x + y = 4t into the equation below?...?... 8y² + 4y = -4x + 6 8y² + 4y = -4(4t - y) + 6 8y² + 4y = -16t + 4y + 6 8y² = -16t + 6 y = sqrt(-2t + 6/8) dy/dt = 0.5 x -2 / sqrt(-2t + 6/8) = - 1/sqrt(-2t +...