How to apply velocity + acceleration to a position?

Question

Steven 2014-11-26

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https://ww2.mathworks.cn/matlabcentral/answers/164329-how-to-apply-velocity-acceleration-to-a-position

编辑： Steven 2014-11-27

采纳的回答： Youssef Khmou

Thank you all.

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Youssef Khmou 2014-11-27

what is the initial position x0,y0?

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Answer 1

Youssef Khmou 2014-11-27

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编辑：Youssef Khmou 2014-11-27

@Roger gave the solution (Vx,Vy) . try to write a feedback of this solution.

 t=0:100e-3:20;
V0x=1000;
Alpha=0.0004;
Beta=0.25;
Vx=1./(Alpha*t+(1/V0x));
Vy=(10/Beta)*(exp(-Beta*t)-1);
x0=10;
y0=15;
x=x0+(1/Alpha)*(log(V0x*Alpha*t+1));
y=(10/Beta)*(-(exp(-Beta*t)/Beta)-t)+y0+(10/Beta^2);
figure; plot(x,y)
title(' Particle Trajectory')
xlabel('x');
ylabel('y');

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Answer 2

Roger Stafford 2014-11-26

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You can approach this problem two ways. One is symbolic and other is numeric. As you are probably aware, you have two entirely independent differential equations here which simplifies things both for the numeric and symbolic methods.

For the symbolic approach you can either use matlab's 'dsolve' function to obtain analytic expressions for x and y versus time t, or you can use your calculus to solve these differential equations by hand. The latter is simple to do. For example, your equation

dvx/dt = -0.0004*vx ^2

can be expressed as

=1/vx^2*dvx = 0.0004*dt

and both sides can easily be integrated.

For the numeric approach you can set up these differential equations to be solved using one of the 'ode' functions. Read about them at:

http://www.mathworks.com/help/matlab/math/ordinary-differential-equations.html

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Answer 3

Youssef Khmou 2014-11-26

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You can verify this primary solution theoretically :

   t=0:100e-3:20;
   V0x=1000;
   Alpha=0.0004;
   Beta=0.25;
   Vx=1./(Alpha*t-V0x);
   Vy=exp(-Beta*t)+10/Beta;

If it is correct, you can integrate for second time to get (x,y)

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Steven 2014-11-26

编辑：Steven 2014-11-27

Thank you for your help this far. I greatly appreciate it. I'm wondering if my values are the same as yours regarding integrals:

Vxi = 2500*log*(0.0004t-1000),

and

Vyi = -40*t-160 * e^(-0.25t)

When I plot Vxi, the integral of Vx it doesn't seem to start at t=0, is there any reason for this? Vyi seems fine, as it starts the graph at t=0. Is there any way I can make the plot of Vxi start at t=0?

Roger Stafford 2014-11-27

I assume that the symbol 'Vxi' means the same as 'x'. If so, I don't quite agree with your result.

What we have already obtained is the equation

vx = dx/dt = 1/(0.0004*t+0.001)

as the result of the first integration. To find x as a function of t, we need to integrate the expression on the right hand side. Its integral is:

x = 1/0.0004*log(0.0004*t+0.001) + C

where C is the appropriate constant of integration. If you want x to be zero when t is zero, then C must be -1/0.0004*log(0.001), which then gives the final answer of:

   x = 1/0.0004*log(0.0004*t+0.001) - 1/0.0004*log(0.001)
     = 1/0.0004*(log(0.0004*t+0.001)-log(0.001))
     = 1/0.0004*log((0.0004*t+0.001)/0.001)
     = 2500*log(0.4*t+1)

Your expression for 'y' ('Vyi') looks basically correct except that it is equal to -160 when t is zero. It needs to have a constant of integration of 160 added if you want it to be zero when t is zero.

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