1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Position vs. Time / Velocity

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    A motion detector transmits a click which is reflected off of a moving car. The echo is received by the motion detector after .0026 seconds. the motion detector sends out a second click .1 seconds after the first click. The echo for the seconds click is received after .0031 seconds.

    1) How far has the car moved?
    2) What is the average velocity?

    2. Relevant equations

    v = Vo + at
    S = 1/2 (Vo+V)t
    U = s/t
    S = vt

    3. The attempt at a solution
    .0026 - .1 = .0974 seconds
    .0031
    .0974 + .0031 = .1005 seconds
    340 (Spd of sound m/s) * .1005 = 34.17 meters ??
    Velocity ??
     
  2. jcsd
  3. Sep 17, 2011 #2
    Are you sure that Doppler Effect does not take place Here??
     
  4. Sep 17, 2011 #3
    My son and I have posted this question because we have spent all day trying to solve it. The question comes from a basic lab project that deals with displacement and velocity. doppler has not been introduced. I have typed in all the info from the question. The only additional piece to add is that the speed of sound is 340 m/s.
     
  5. Sep 17, 2011 #4
    Your question can be easily solved by assuming coordinate system(I hope that coordinate axis have been introduced). Assume motion detector lying along X-axis and signal transmitter lying at origin(0,0). Assume car traveling along line y=1(see attachment). Take 2 positions along this line as in your question.Let 1 st position(posn in attachment) be of coordinates (x,1) & second position be (y,1). Calculate distance between 1 and origin using distance formula.Let this be m.
    Now (2m/t1)=v
    Where t1 is time for first echo(given) and v=velocity of sound.
    Similarly calculate distance between second point and origin and let this be n.
    Now (2n/t2)=v
    Where t2 is time for second echo(given) and v=velocity of sound.

    Using 2 equations above calculate x and y individually and again apply distance formula between points (x,1) and (y,1) where we have now found out x & y.

    PS:I am no great hands at Drawing and i accept this diagram is filthy.But i hope you understand.
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook