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Homework Help: Motion involving speed cameras

  1. Jan 1, 2018 #1
    1. The problem statement, all variables and given/known data
    task.png task b.png
    2. Relevant equations

    Equations of motion:

    v = u + at
    s = 1/2 (u + v) t
    s = ut + 1/2at^2
    v^2 = u^2 + 2as

    3. The attempt at a solution

    Hello, on a i used the 2nd equation:

    30m = 1/2 (3*10^8m/s+3*10^8m/s) t so -> t = 1*10^-7s

    but to get the correct solution i need to take t*2 i think. I made a small sketch to this question, but i dont really know why i need to take it times 2, is it because of sending and receiving the signal?
    I hope someone can explain this to me.


    b was a lot more confusion for me because of the two time values:

    first i used s = v*t

    s = 3*10^8m/s * 2.14 * 10^-7s so -> s = 64.2m

    this is where i get lost, so now i used the 2nd equation again with the other time value:

    64.2m = 1/2 (0m/s + final velocity) * 0.1s so -> v (final velocity) = 1284 m/s

    so im pretty sure thats waaay to high, but i think the first part is correct.
    Thanks for every answer.
     
  2. jcsd
  3. Jan 1, 2018 #2

    PeroK

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    Exactly, the light pulse must travel to the car, be reflected and travel back again.

    You need to think more clearly what happens in the ##0.1s## between pulses. What is different? Does anything move ##64.2m##?
     
  4. Jan 1, 2018 #3
    thanks for your answer,

    i think only the car moves further away in 0.10s but i dont have the velocity of the car, so i cant figure out how far it gets, can i? Or do i need to subtract the 0.1s from the given 2.14*10^-7s to make up for the delay of the 2nd pulse?
     
  5. Jan 1, 2018 #4

    PeroK

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    Finding the speed of the car is the point of the speed gun!

    Fundamentally, what does firing a pulse of light at a car and measuring the time it takes to come back tell you? If you fired a pulse of light at the moon and it reflected and came back ##2.5s## later, what does that tell you about the moon?
     
  6. Jan 1, 2018 #5
    it tells us the distance to the moon, doesnt it? That means we can use the time that the pulse travels and its known velocity to determine the distance to the car (64.2m). With the following pulse we can then determine the distance the car moved in the next 0.1s, so the total distance over the time both pulses took to reach it, is the cars velocity, is that correct?
     
  7. Jan 1, 2018 #6

    PeroK

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    That's the idea. But, be careful about how far the pulse travels (hint: the pulse makes a return trip).

    Hint: what happens if the car is not moving?
     
  8. Jan 1, 2018 #7
    All right now i think i got it. The distance it was away after 0.1s wasn´t 64.2m but 32.1m, i got confused because of the return trip.
    So the change in distance in 0.1s is 32.1m - 30m = 2.1m and now we know the distance the car moved in 0.1s, so:

    v = 2.1m/0.1s = 21m/s -> 75.6 km/h

    is that correct?
     
  9. Jan 1, 2018 #8
    Krashy,
    If you multiply the speed of light by the time between pulse emission and registration, won't that give about twice the
    final distance to the car? Thanks for posting this question!
     
  10. Jan 1, 2018 #9
    Yeah that´s why you halve the time between pulse emission and registration, so you only get the distance from emission to the back of the car. So in this case for b you dont use 2.14*10^-7s but 1.07*10^-7s to get the right value.
     
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