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First year practise problem

  1. Nov 14, 2013 #1
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    3. The attempt at a solution
    I tried solving this question a few different ways, and I got different answers every time. My first attempt I realized that the relationship between the changing x, changing y, and changing angle was tan([itex]\theta[/itex]) = y/x , so first of all I tried to figure out the final x, and I could do this because it gave me the angular velocity and the final angle, so I got the time to be ∏/2. Then I multiplied 2*(∏/2) to get the distance travelled. So the distance from the wall is 10 - ∏, now that I know the x I can figure out dy/dt, so I differentiate the first expression and get (sec([itex]\theta[/itex]))^2*d[itex]\theta[/itex]/dt = dy/dt * 1/(10-∏). Then the answer would be (sqrt(2))^2 * 0.5 * (10-∏) = dy/dt, so that was my first answer, now here to my second answer. I differentiate that first expression straight from the beginning, I won't post it but I used the quotient rule, then I figured out that I needed x and y, so I figured out the time the same way as before, it was ∏/2, then I figured out the distance from the wall which was also 10 - ∏, but I also figured out that tan(∏/4) = 1 = O/A so x = y, so y = 10 - ∏ also, but when I plug everything in then, I get 8.86 m/s instead of 6.86 m/s which was what I got in the first attempt. But both of these answers are wrong according to the prof.
     
  2. jcsd
  3. Nov 14, 2013 #2

    Dick

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    You've got some parts right. y=x*tan(θ). Differentiate both sides with respect to t and remember x AND θ are both functions of t. Use the product rule.
     
  4. Nov 14, 2013 #3
    dy/dt = tan(θ) *dx/dt + x*(sec(θ))^2*dθ/dt
     
  5. Nov 14, 2013 #4

    Dick

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    Ok, now use that.
     
  6. Nov 14, 2013 #5
    dy/dt = 2 + (10 - ∏)*2*0.5
    = 8.86 m/s
    but my professor got 4.86 m/s...he used -2 instead of 2....sooo is he right?
     
  7. Nov 14, 2013 #6

    Dick

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    Yes, he is. dx/dt isn't 2. x is getting smaller as time goes on.
     
  8. Nov 14, 2013 #7
    ohh alright, thanks!
     
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