# Am I doing this right?

4 Ansichten (letzte 30 Tage)
Jacob Savona am 23 Feb. 2015
Bearbeitet: Rick Rosson am 23 Feb. 2015
You are driving a car along a straight road for t = 50s. Assume that friction is negligible but drag is not, where F .5b(v_n-1)^2 , b is a given constant, and v_n-1 is the velocity at the previous time step. For the duration of the trip as described below, calculate the position, velocity, and netacceleration of the car. Use one or more for loops to simulate the behavior of the car starting at t = 0 until t = 50s, with Δ t = 0.5s. Mass of car and you=1428.8kg and b= 1.04 N*s2/m2.
a. Starting from rest, where initial position of x(0) = 0 and v(0) = 0, you apply an initial, constant acceleration of a = 3m/s , until you reach 65 mph(29.05m/s). Upon reaching 65 mph, you maintain a constant velocity.
Here's My code:
m=1428.8;
x=0;
a=3;
for t=[0:.5:50]
v_n=a.*t;
F_drag=.5*(1.04)*v_n;
F=m*a
F_net=F-(F_drag);
if F_net==0
v_n;
p=v_n/a;
end
end
##### 1 Kommentar-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
Jacob Savona am 23 Feb. 2015
Fdrag = .5b(v_n-1)^2 for any confusion

Melden Sie sich an, um zu kommentieren.

### Antworten (2)

Rick Rosson am 23 Feb. 2015
Bearbeitet: Rick Rosson am 23 Feb. 2015
% Mass of the vehicle (in kilograms):
m = 1428.8;
% Applied acceleration (in m/s^2):
a_gas = 3;
Time increment (in seconds):
dt = 0.5;
% Time vector (in seconds):
t = (0:dt:50)';
% Number of data samples:
N = length(t);
% Pre-allocate state variables:
x = nan(N,1);
v = nan(N,1);
a = nan(N,1);
% Initial conditions:
x(1) = 0;
v(1) = 0;
a(1) = a_gas;
% Main simulation update loop:
for n = 2:N
...
...
a(n) = ...
v(n) = v(n-1) + ...
x(n) = x(n-1) + ...
...
...
end
##### 6 Kommentare4 ältere Kommentare anzeigen4 ältere Kommentare ausblenden
Rick Rosson am 23 Feb. 2015
Bearbeitet: Rick Rosson am 23 Feb. 2015
I don't understand - you already know how to calculate the drag, you included the calculation in the code that you posted as part of your question.
You have everything you need to solve this problem. Try something in MATLAB, check to see if it works, and if it does not, then try to figure out how to fix it. If you want to learn physics and/or MATLAB, you have to put in some effort. No one is going to do it for you.
Best of luck.
Rick Rosson am 23 Feb. 2015
BTW, the drag is NOT a function of time. It's a function of velocity. So if you first calculate the velocity v at time step n, then you can calculate the drag. Once you know the drag, you can calculate the net force. And then the acceleration....

Melden Sie sich an, um zu kommentieren.

Image Analyst am 23 Feb. 2015
##### 3 Kommentare1 älteren Kommentar anzeigen1 älteren Kommentar ausblenden
Image Analyst am 23 Feb. 2015
Drag is a force in the opposite direction to acceleration. Remember Netwon's law, F=m*a. So you have -Drag/m as a negative acceleration, so add that to a, which will make a smaller. Unless your professor gave you some other formula...
Rick Rosson am 23 Feb. 2015
Bearbeitet: Rick Rosson am 23 Feb. 2015
Not quite correct. Drag is a force in the opposite direction to velocity, not acceleration. The net acceleration may or may not be in the same direction as the velocity. And the component of the acceleration resulting from the drag will always point in the same direction as the component of the force resulting from the drag (as required by Newton's second law).

Melden Sie sich an, um zu kommentieren.

### Kategorien

Mehr zu Programming finden Sie in Help Center und File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by