
#1
Dec1309, 06:21 PM

P: 1

1. The problem statement, all variables and given/known data
Calculate the position (distance) and velocity of the plane at 5 equal time intervals based on constant acceleration until the takeoff point. The 5th point should be the takeoff point. Calculate the abort point, if the plane needs to make a sudden stop. Determine the distance to the final stopping point after landing. FBD: Calculate all of the forces acting on the plane while it is flying at a constant velocity and altitude. Graphs: Graph velocity of the plane versus time and distance versus time for the 8 time intervals (where in the problem can you infer that there are 8 points?) Here's what information I have: Mass of airplane: 1060 kg Takeoff speed: 65 knots> 33.44 m/s Engine Thrust: 1500 N Landing speed: 72 knots> 37.04 m/s Maximum abort speed: 58 knots> 29.834 m/s FAA noise abatement height for takeoff: 150 m Breaking coefficient of friction: 0.30 Take off angle: 35 degrees Runway Design Terms: Abort velocity The FAA establishes a point during take off when the pilot must shut down and come to a stop or if the pilot passes this point he/she must take off. The runway must include enough length for the pilot to use reverse engine thrust and braking friction to stop safely if the decision is made not to take off. Takeoff angle The maximum angle to the horizontal that the pilot can use to gain altitude just after take off. FAA Noise Abatement The FAA sets minimum flights standards for planes flying over populated areas. These minimum heights must be met by placing runways far enough from residential areas to allow the planes to meet these standards. Planes normally maintain their take off speeds until the exceed these minimum heights. 2. Relevant equations a=[tex]\Delta[/tex]v/t vf=vi+at d=vit+1/2at^2 v=d/t 3. The attempt at a solution *Units included This is my attempt at finding the time: a=[tex]\Delta[/tex]v/t 9.8m/s/s=3.6/t t=.3673 s This is my attempt at finding distance: d=vit+1/2at^2 d=(33.44m/s)(.3673s)+1/2(9.8m/s/s)(.3673)^2 d=12.94 m This is my attempt at finding a second distance: d=(33.44m/s)(.7346s)+1/2(9.8m/s/s)(.7346s)^2 d=27.21m  This is for my Unit project. Everything in the given section, is what I have been given. My current grade is a 73%. I try my ABSOLUTE hardest. I do perfectly fine on classwork and homework but bomb tests and quizzes why, I have NO idea whatsoever. I really would like to get a good grade on this project. I have calculated 5 distances but I included two because I just want to know if I am on the right track. I am making an assumption right now that this website is going to become my lifesaver you truly don't understand how frustrating getting the type of grades I get is for me, when I try my absolute hardest. It's almost like I do everything for nothing. I try so so hard. I want to do good. I am graduating high school a year early, and physics is bringing my GPA down and it is frustrating me so much. Please please please and thank you thank you thank you in advance! P.S. I'm new here and this is my first post, so if something is missing or whatever, don't hurt me, I'm still learning :) 



#2
Dec1309, 07:16 PM

P: 1,351

I can see why it is frustrating. It is a great question,, however.
Thankfully it doesn't involve lift which might do what to the frictional force between the wheels and aircraft? Ok not mentioned, so just consider the result of ma=Force of acceration  frictional force. Here friction force is equal to the weight of the aircrafttimes the coefficient of friction. So you can calulate net force and acceleration from that, and can solve the problem. Recall V=a*t 


Register to reply 
Related Discussions  
Airplane design  Mechanical Engineering  7  
Airplane problem. 2d3d motion  Introductory Physics Homework  16  
Airplane drops a package (relative motion)  Introductory Physics Homework  11  
uniform circular motion  airplane  Introductory Physics Homework  3  
circular motion of airplane  General Physics  5 