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Simple trig problem has me stumped

  1. Jun 5, 2008 #1
    Hi

    I'm sure this is an easy one, but I've managed to thoroughly confuse myself. Basically I'm trying to come up with a formula to determine how long a signal takes to travel from point a to point b.

    Here are the givens:

    a and b lie on parallel lines where the distance between the lines is d and a line drawn from a to b is perpendicular to both lines.

    if a and b are at rest, the time it takes a signal to go from a to b is t

    if a and b are moving in the same direction at speed t/m, how long would it take a signal from a to reach b?

    This seems like a simple right triangle relationship but I can't seem to figure it out.

    Any help appreciated.


    P.S. This isn't homework. It's for part of a Doppler shift program I'm writing.
     
    Last edited: Jun 5, 2008
  2. jcsd
  3. Jun 5, 2008 #2

    HallsofIvy

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    What kind of "signal" are you talking about? If this is a electromagnetic signal (e.g. light or radio) then, since the everything can be done relative to a coordinate frame in which one (and therefore both, since they move in the same direction at the same speed) is stationary, the time would be t, exactly the same as when both are at rest (which they are in this frame).

    If the signal is a sound wave, in a medium in which the speed of sound is v0= d/t, then set up a coordinate system in which the origin of the signal is at (0,0) and the receiver is initially at (0, d). If they are both moving with velocity u, then at time T, the receiver will be at (uT,d). In order that the signal be received then, we must have the distance from (0, 0) to (uT, d) equal to v0T= dT/t. That is [itex]\sqrt{u^2T^2+ d^2}= dT/t[/itex] or [itex]u^2T^2+ d^2= d^2T^2/t^2[/itex] or [itex]((d^2/t^2)- u^2)T^2= d^2[/itex]. Solve that for T.
     
  4. Jun 5, 2008 #3
    The signal is abstract. Could be a tennis ball shot from air gun, an image of a clock, a yak with a jet pack, whatever. The second formula was what I was looking for. I plotted it out in Adobe Illustrator and it works perfectly.

    Thanks HallsOfIvy
     
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