Recent content by AirForceOne

Okay, I understand now. So A must be diagonalizable in order to compute its exponential. For A to be diagonalizable, its eigenvectors must be linearly independent. I calculated A's eigenvalues to be σ (repeated). When I calculated its eigenvectors, I get [anything, anything]. Does this translate...

Hi, Suppose I have this matrix A: Why is the matrix exponential like so: when A is not simple (eigenvalues not distinct)? Thanks.

But that's what I did... y=(1/2)gt^2 t=sqrt(2y/g) =sqrt((2)(2.5)/9.8) = 0.71 0.71-0.25 = .46... Diagonal distance=2.5, y=(1/2)at^2 a=(2y)/(t^2) =(2*2.5)/(0.46^2) = 23.63ft/s^2...which is what I got before.

Homework Statement A person trips against a table, causing a glass to fall off the edge. An excellent human reaction time is 0.25 seconds. In that time, how far wil the glass fall? Assume that immediately (after 0.25 s has elapsed) begin to acclerate your hand so that you grab the glass when...

Thank you. I was able to find part a) and r1. However, I was unable to find r^.. in part b). This is what I have so far: I don't know how to integrate the acceleration vector to velocity vector it seems.

Hi, Say I have an acceleration vector in polar coordinates: a = -30e_r where the unit vector e_r points in the same direction as the Cartesian unit vector j. How can I integrate that vector so that I have the velocity vector in polar coordinates? I know that if I have an acceleration vector...

Homework Statement Homework Equations The Attempt at a Solution But the correct answer is 1227 ft/s^2. Thanks.

Thank you very much everyone. I understand it now.

Thank you very much. I got 0.06. What is the meaning of the second part of the question ("Does the acceleration of gravity make any contribution")? Thanks again.

Homework Statement A person throws out the back of a car a strong magnet that is attracted to the metal of the car. Will the magnet return and stick to the back of the car or fall to the ground? The initial velocity of the magnet is v=sqrt(5)i m/s. Treat the magnet/car interaction as an...

Why isn't it 3km?

For the graphs, I used Powerpoint. It took a long time though. Next time I'll probably use Solidworks or Autocad, which is a complete overkill but it'll get the job done much, much quicker. For the text, I used MathType. I agree it is a simple kinematics problem, but I was just offering another...

Because you actually looked up the length of a Jackson's chameleon, I created a lizard just for you to applaud your enthusiastic attitude:

Yeah it's 0.20m. Made an calculation error :P. Because I am such an awesome guy (and still waiting for my question to be answered), I decided to graphically show how I did this problem without using the kinematic equations (I hate those equations).

Hi, While I'm waiting for my question to be answered, I thought I'd help you out. Is the answer 0.25 meters? I might be completely off, but like I said, just trying to help haha.