Homework 03 Posted by D. Dane Quinn on Fri 30 of Jan, 2009 12:03 EST Another week...more snow...more homework.

Homework 03 Posted by D. Dane Quinn on Fri 30 of Jan, 2009 12:03 EST Another week...more snow...more homework.

Posted by ChrisLane on Mon 02 of Feb, 2009 16:25 EST I don't know if anyone has encountered this, or if I am just doing something completely wrong, but I keep getting the velocity to be 18.976 m/s. The group that I work with has done this several different ways, and still come out with a different answer than in the back of the book. Is it possible that the back of the book is wrong, or am I just looking over something? We have been using the equations of motion with constant acceleration.

Posted by DominickDeMasiJr on Mon 02 of Feb, 2009 17:30 EST NUMBER 1. Im not sure if I'm integrating a=(6+.02s) correctly. I took it as v=(6+.02s)t^2 and ect. but when I get position plug in 100 for s and solve for t I get 5 not 5.62. NUMBER 2 I found all my equations of motion, and just can't see how to set them up to solve for t. I could understand how you could treat them as a piecewise function, and I've heard people say you need to use integration.

Posted by ChrisArnold on Mon 02 of Feb, 2009 20:32 EST You are using the acceleration due to gravity when in meters (9.8) but the units in this problem are feet so you should use 32.2 for standard gravity.

Posted by D. Dane Quinn on Mon 02 of Feb, 2009 22:28 EST Problem 01The acceleration is given as a function of position, so you cannot simply integrate with respect to time. There is a similar example in the book, but unfortunately I'm home now and don't have the textbook with me. Any help out there? Problem 02Yes, your peeps are right-you will have to integrate. The values of the acceleration are known, but the time that the car switches from acceleration to deceleration is unknown, as is the final time. You will have to perform the integrations including these unknown times and then solve for them using the final velocity and position of the car.

Posted by CameronClose on Mon 02 of Feb, 2009 22:56 EST Number 7 seems too easy? Are people getting Ar as -.25, and A theta as -3.2?

Posted by DavidSeveryn on Tue 03 of Feb, 2009 00:30 EST > Problem 01The acceleration is given as a function of position, so you cannot simply integrate with respect to time. There is a similar example in the book, but unfortunately I'm home now and don't have the textbook with me. Any help out there? > > Problem 02Yes, your peeps are right-you will have to integrate. The values of the acceleration are known, but the time that the car switches from acceleration to deceleration is unknown, as is the final time. You will have to perform the integrations including these unknown times and then solve for them using the final velocity and position of the car. For Problem 1, the example you are referring to is Example 12.4 on page 13. It helps...but I still can't get the right answer...probably doing something wrong because I am ending up with a pretty difficult integral that I had to use Maple to solve...then I got a solution of like 70 seconds so yeah that's wrong.

Posted by MikeBenedick on Tue 03 of Feb, 2009 09:21 EST Problem 1: Use example 12.4 p13 to get a similar integral of ds/...?, and then refer to Appendix A table of integrals p657 in order to plug-in the necessary values. I got an answer of 5.625m.

Posted by ChrisLane on Tue 03 of Feb, 2009 10:50 EST Dave, that integral you get is solvable, it is just really difficult.I ended up getting the integral of t dt equals the integral of 1/(12s+.02s*2)*(1/2)ds with bounds from 0 to 100, which gives t then to be 5.62 secs. p.s. i used asterisks because the carrots wouldn't display, those are powers.

Posted by DavidSeveryn on Tue 03 of Feb, 2009 11:18 EST > Dave, that integral you get is solvable, it is just really difficult.I ended up getting the integral of t dt equals the integral of 1/(12s+.02s*2)*(1/2)ds with bounds from 0 to 100, which gives t then to be 5.62 secs. > > p.s. i used asterisks because the carrots wouldn't display, those are powers. Yeah that's what I got...you'll have to show me how...I remember doing something like that before but I was drawing a blank last night.

Posted by DominickDeMasiJr on Tue 03 of Feb, 2009 12:43 EST Ok, I got though most of this, set up the t= integral 1/(v) ds.... looked at the appendix found the closest fit to what we have. plug n chugged for 100 and got 32.124ect. which works out to t=8.013 I have checked it several times it doesn't look like I had an error in my numbers, but I know I should be getting 15.792 where I'm getting 32. Only thing I can figure is I'm using the wrong equation in the appendix?? I used the third one down on the right hand side of the page and just took a=0. I've tried it for both solutions given. The Ln gave me the 32. and sin-1 gives me 5.6 and 11.1

Posted by AngieTriplett on Tue 03 of Feb, 2009 14:53 EST Yes, I know the integral is slightly ugly, but you can use matlab to solve it or using an integral table of the form int(1/sqrt(a*u+u*u)) du = ln( sqrt(a*u + u*u) + u + a/2 ). Just let me know which method you use.