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I proved a physics equation. Does any one know if theirs an equation like this?

  1. Oct 19, 2014 #1
    Direction of a plane proof
    48. A pilot wants to land at a small lake that is [N30.0 ̊W] of the airport that she is starting from. The wind has a velocity of 25.0 m/s[W] and the air speed of the plane is 1.90 × 10^2 m/s . What direction will the plane have to fly to get to its destination? What will be the velocity of the plane relative to the ground?

    vt as the speed you'll achieve .
    x as the plane speed or velocity.
    w as wind speed or velocity.
    Θ as the angle the plane has to go to achieve angle Φ at vt speed.
    Φ as the angle away from start.










    Know start to take the big picture coloured in red
    vt^2=(xsinΘ+w)^2+(xcosΘ)^2
    vt^2=(x^2sinΘ^2+wxsinΘ+wxsinΘ+w^2)+(x^2cosΘ^2)

    now we have to get rid of the cosΘ^2 by taking(in green)
    x^2=(x^2sinΘ^2)+(x^2cosΘ^2)
    With x^2 divided
    x^2/x^2=(x^2sinΘ^2)/x^2+(x^2cosΘ^2)/x^2
    1=sinΘ^2+cosΘ^2
    1-cosΘ^2=sinΘ^2
    Now sub in to the red coloured equation

    vt^2=x^2(1-cosΘ^2)+wxsinΘ+wxsinΘ+w^2)+(x^2cosΘ^2)
    vt^2=(x^2-x^2cosΘ^2)+(wxsinΘ+wxsinΘ+w^2)+(x^2cosΘ^2)
    vt^2=x^2+2wxsinΘ^2+w^2

    Now look at what's coloured in blue which is used to get rid of vt
    sinΦ=(xsinΘ+w)/vt
    (vt=(xsinΘ+w)/sinΦ)^2
    vt^2=(x^2sinΘ^2+wxsinΘ+wxsinΘ+w^2)/sinΦ^2

    Now we sub this equation into the red and green equation

    (x^2sinΘ^2+2wxsinΘ+w^2)/sinΦ^2=x^2+2wxsinΘ+w^2

    Move everything over to left

    (x^2sinΘ^2/sinΦ^2)+(2wxsinΘ/sinΦ^2-2wxsinΘ)
    (+w^2sinΦ^2-x^2-w^2)=0
    A B C
    Use the quadratic formula
    sinΘ=-(2wx/sinΦ^2-2wx)+/-square root((2wx/sinΦ^2-2wx)^2-4x^2/sinΦ^2(w^2/sinΦ^2-x^2-w^2)/2x^2/sinΦ^2

    sinΘ=(-2wx/sinΦ^2+2wx)+/-square root((4x^2w^2/sinΦ^4-4w^2x^2/sinΦ^2-4w^2x^2/sinΦ^2+4w^x^2)(-4x^2w^2/sinΦ^4+4x^4/sinΦ^2+4w^2x^2/sinΦ^2))/2x^2/sinΦ^2

    sinΘ=(-2wx/sinΦ^2+2wx)+/-square root((-4w^2x^2/sinΦ^2+4w^x^2+4x^4/sinΦ^2))/2x^2/sinΦ^2

    sinΘ=(-2wx/sinΦ^2+2wx)+/-square root((-4w^2x^2+4x^4)/sinΦ^2+4w^2x^2)/2x^2/sinΦ^2

    multiply the whole equation by sinΦ^2
    sinΘ=(-2wx+2wxsinΦ^2)+/-square root((-4w^2x^2+4x^4)sinΦ^2+4w^2x^2sinΦ^4)/2x^2
    divid the equation by 2

    sinΘ=(-wx+wxsinΦ^2)+/-
    square root((-w^2x^2+x^4)sinΦ^2+w^2x^2sinΦ^4)/x^2
    Divid by x
    sinΘ=(-w+wsinΦ^2)+/-square root((-w^2+x^2)sinΦ^2+w^2sinΦ^4)/x

    now say + not +/- and factor out sinΦ out of what's in the
    Square root since we don't want a negative angle.

    sinΘ=(-w+wsinΦ^2)+square root((-w^2+x^2+w^2sinΦ^2))sinΦ/x
    Also written as

    sinΘ=square root((-w^2+x^2+w^2sinΦ^2))sinΦ+(-w+wsinΦ^2)/x
    if wind is in y direction
    cosΘ=square root((-w^2+x^2+w^2cosΦ^2))cosΦ+(-w+wcosΦ^2)/x

    x>w
    If wind is going to opposite your going make negative.
     
  2. jcsd
  3. Oct 19, 2014 #2

    phinds

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    Just as an aside on your problem, addressing your terminology. You cannot "prove" anything in physics, only in math, so if you proved it, it's math. Physics is always subject to disproof, math not so much.
     
  4. Oct 19, 2014 #3
    Thank you for the advice I'm only in grade 11 and I didn't take physics yet
     
  5. Oct 19, 2014 #4

    phinds

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    Well, welcome to the forum
     
  6. Oct 19, 2014 #5
    Hey thanks and this is my first time posting so that's why I didn't know what section to put my formula question in and by the way this is only for x direction the other one for cos I proved mathematically on paper but didn't type up yet.
     
  7. Oct 19, 2014 #6

    russ_watters

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    Staff: Mentor

  8. Oct 20, 2014 #7
    It maybe long but I did so I can create an equation to solve problems like this since I like using equation and wouldn't sin and cosine laws work too
     
  9. Oct 21, 2014 #8

    CWatters

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