Computer science student needing help in solving an exercise

  • Thread starter erkant
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  • #1
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I even don't know if this is the appropriate place to post this questions, while as long as I know this place is considered for research topics and more advanced topics in physics. The point is that I need to write a MATLAB code for specific question, but first of all I need to solve it, and show the step by step solution of it, and then code it. Because I am not that good in physics, I though maybe someone where can give me a step by step solution, so I can code it later.

Here is the question:

DNl4wVJ.jpg
 

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  • #2
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Well, if you want to calculate the EXACT velocity, you would need to know the angles alpha and beta as a continuous function of time, in other words, you would need to know their values at ALL times. Since this is not provided to you, but rather only their values at specific instants of time, you would only be able to arrive at an APPROXIMATION. The broad way of doing this would be:
x component of velocity, vx = dx/dt
y component of velocity, vy = dy/dt
magnitude of velocity, v = sqrt(vx^2 + vy^2)
angle gamma = tan -1 (vy/vx)
Remember from your table that you don't know d(alpha)/dt and d(beta)/dt so you would have to use a suitable approximation technique to arrive at their approximate values at time t = 10, from the data provided in the table. One way to do this would be to replace d(alpha)/dt by (delta)(alpha)/(delta t), in other words use average values instead of instantaneous values.
 
  • #3
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Well, as I said I need to program it, therefore I need the full path to the solution. I don't know that much physics. In programing instead of degrees, radians are used, so I wrote a program that converted those degrees to radians and calculated x and y.

I got the following results: x=1531.04, y=2170.38
I need only to calculate the speed v and climb angle gamma at t=10. So how, should I continue?
 
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  • #4
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Just differentiate the expressions for x and y w.r.t. time and then plug in the values of d alpha/dt and d beta/ dt. C'mon dude, do I have to spell out every step?The only physics you need to know really is that velocity is the rate of displacement. The rest is just math. And if I am not mistaken a calculus based physics course is a prerequisite even for computer science students in most universities. And during my ex-TA days I've even observed that some of them even fare better than physics students, so please don't give me that excuse. Cheers.
 
  • #5
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I understood from your earlier post what I should do (or at least I hope so). For example, vx = dx/dt, how can I differentiate with respect to t, when there is no t in the function. That's my problem, and question. Should I use something like implicit differentiation?
 
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  • #6
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I understood from your earlier post what I should do (or at least I hope so). For example, vx = dx/dt, how can I differentiate with respect to t, when there is no t in the function. That's my problem, and question. Should I use something like implicit differentiation?
The best you can do here is to use the angles given for each time to compute an (x,y) coordinate

and then to use the points computes to get vx and vy

vx = x(10) - x(9)
vy = y(10) - y(9)

and similarly for the other velocity point.

That would give you two velocity points the one between t=9 and t=10 and the one between t=10 and t=11.

You can then find the average velocity of the plane.
 
  • #7
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And what if I calculate the average? Is there any way that I can calculate the velocity at t=10?

If I go like you say:

vx = x(10) - x(9) = 1450.50 - 1401.92 = 48.58
vy = x(10) - x(9) = 2000.84 - 1987.35 = 13.49

v = sqrt(vx^2 + vy^2) = sqrt(48.58^2 + 13.48^2) = sqrt(2541.72) = 50.41

vx = x(11) - x(10) = 1450.50 - 1498.64 = -48.14
vy = x(11) - x(10) = 2000.84 - 2013.51 = -12.67

v = sqrt(vx^2 + vy^2) = sqrt(-48.14^2 + -12.67^2) = sqrt(2477.98) = 49.77

avgv = 50.41 + 49.77 / 2 = 50.09

do you mean that this is the velocity at time t=10?
 
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  • #8
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when you average the two velocities the result is the avg velocity at t=10
 
  • #9
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vx = x(10) - x(9) = 1450.50 - 1401.92 = 48.58
vy = x(10) - x(9) = 2000.84 - 1987.35 = 13.49

v = sqrt(vx^2 + vy^2) = sqrt(48.58^2 + 13.48^2) = sqrt(2541.72) = 50.41

vx = x(11) - x(10) = 1450.50 - 1498.64 = -48.14
vy = x(11) - x(10) = 2000.84 - 2013.51 = -12.67

v = sqrt(vx^2 + vy^2) = sqrt(-48.14^2 + -12.67^2) = sqrt(2477.98) = 49.77

avgv = 50.41 + 49.77 / 2 = 50.09

do you mean that this is the velocity at time t=10? please note that degrees are converted to radians and x and y are calculated like this for the sake of my exercise.
 
  • #10
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vx = x(10) - x(9) = 1450.50 - 1401.92 = 48.58
vy = x(10) - x(9) = 2000.84 - 1987.35 = 13.49

v = sqrt(vx^2 + vy^2) = sqrt(48.58^2 + 13.48^2) = sqrt(2541.72) = 50.41

vx = x(11) - x(10) = 1450.50 - 1498.64 = -48.14
vy = x(11) - x(10) = 2000.84 - 2013.51 = -12.67

v = sqrt(vx^2 + vy^2) = sqrt(-48.14^2 + -12.67^2) = sqrt(2477.98) = 49.77

avgv = 50.41 + 49.77 / 2 = 50.09

do you mean that this is the velocity at time t=10? please note that degrees are converted to radians and x and y are calculated like this for the sake of my exercise.
Yes, that looks right (I didn't verify your calculations but the concept looks right).
 
  • #11
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Time is not explicitly present in the equation, but you know that alpha and beta are functions of time. Hence, you can get dx/dt as a function of alpha, beta, d (apha)/dt and d(beta)/dt. Then try to plug in the approximate values of d (apha)/dt and d(beta)/dt to get the approximate value of dx/dt. What jedishrfu is suggesting should work too.
 
  • #12
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So assuming my calculations and concept is right. How can I calculate the angle.

If the equation for angle is: gamma = tan -1 (vy/vx), which vy/vx should I use? Or should I also use vy and vx for all times like above and do the average?

@physwizard isn't what you suggest, what I already do? I mean just plug the values of alpha and beta at specific time to expressions of x and y?
 
  • #13
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I think you can use the v average and tan(gamma) = vy/vx

you could try using both velocity vectors but I think it will work out the same.
 
  • #14
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I get around 88 degrees of angle.

gamma = tan -1 (vy/vx) = tan -1 (50.09) = 88 degrees.

According to picture it seems much more small. Is there any problem or I am just being fastidious? Can someone just do the calculations and verify me please?
 
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  • #15
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No, I was suggesting something different but I feel what jedishrfu is suggesting would work out to be simpler. If your previous calculations are correct, vy/vx is around 0.25, so gamma is arctan of that which should be around 14 degrees.
 
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  • #16
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I get around 88 degrees of angle.

gamma = tan -1 (vy/vx) = tan -1 (50.09) = 88 degrees.

According to picture it seems much more small. Is there any problem or I am just being fastidious? Can someone just do the calculations and verify me please?
The 50.9 value is the length of the velocity vector you need to use trig the compute the angle.

Physwizard did it right:

vy / vx = arctqn(gamma)

You have to average the vx and vy values individually and use those values in computing the arctan of gamma.

And 14 degrees seems right not many planes shoot upward at 88 degrees.
 
  • #17
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Okay got it, thanks!
 

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