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Frequency of Trucks Overtaking a Cyclist

  1. Jul 15, 2017 #1
    1. The problem statement, all variables and given/known data

    Garbage Trucks, one after the other, are heading for the dump. They move at 19.7 m/s, and every 3 min, 2 of them arrive at the dump. A cyclist, moves at 4.47 m/s, heading for the dump as well.

    a) With what frequency do the trucks overtake the cyclist?
    b) What if there was a hill, that didn't phaze the trucks, but resulted in the cyclist going at 1.56 m/s. What's the frequency now?

    2. Relevant equations

    v = Δx/Δt

    3. The attempt at a solution

    Well, first of, I decided to find the distance from the starting point of the trucks to the dump.

    vt = Δx/Δt <=> Δx = 19.7 m/s * 180 s = 3546 m

    Using that, I found the time the cyclist needs to cover the same distance.

    vc = Δx/Δt <=> Δt = 3546 m / 4.47 m/s = 793.3 s = 13.22 min

    Then I figured that I'd say:

    -The cyclist needs 13.22 minutes to reach the end.
    -2 trucks need 3 minutes to reach the end.
    -So, how many times do the trucks cover that distance in the time the cyclist does?
    -So: N = 13.22 /3 = 4.41 times

    Now, we have that: In 13.22 minutes, 2 trucks overtake the cyclist, 4.41 times, therefore, 8.82 trucks overtake the cyclist in that timespan. In one minute, how many trucks overtake him?

    f = 8.82/13.22 = 0.667

    And that's where I get stuck, because the book's answer is: 0.515

    Obviously I'm missing something, but I just can't figure out what. As for (b), I'm not sure if the whole drive is a hill, or if it's just a small hill. But it doesn't specify any points or whatnot, so I assume it's the former (trucks keep the same speed all the time, but the cyclist changes his, and it's basically (a) with different numbers).

    Any help is appreciated!
     
  2. jcsd
  3. Jul 15, 2017 #2

    jbriggs444

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    In 13.22 minutes, 2 trucks make it to the dump 4.41 times. Not all of them passed the cyclist though. Some were already on the road.
    How many trucks make it to the dump every minute?
     
  4. Jul 15, 2017 #3

    TSny

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    Looks like your initial setup at t = 0 is
    upload_2017-7-15_12-48-48.png
    Do the gray trucks count as passing the cyclist during the time interval between t = 0 and t = 13.22 min?
    [EDIT: I see jbriggs444 already pointed this out.]

    A nice way to approach this problem is to consider how fast the trucks are moving relative to the cyclist.
     
  5. Jul 15, 2017 #4
    Well, technically the two trucks at the front aren't really overtaking him. In 3 minutes they'll be at the dump, and the cyclist will be at 804.6 m.

    How is that going to help though? I can't grasp that.

    Well:

    2 trucks reach the dump in 3 minutes. | 3x = 2 <=> x = 2/3 trucks
    x trucks reach the dump in 1 minute. |
     
  6. Jul 15, 2017 #5

    TSny

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    So, out of your 8.82 trucks, how many actually pass the cyclist in 13.22 minutes?

    In the cyclists frame of reference, the cyclist considers herself at rest and the trucks are moving relative to her at a certain speed. If you can figure out that relative speed, you can easily find the frequency the trucks pass her.
     
  7. Jul 15, 2017 #6
    So it's 8.82 - 2 (the ones at the front) = 6.82 trucks. And then, f = 6.82/13.22 = 0.516 ? The book's answer is 0.515, so that's probably correct (he keeps 3 Significant Digits, so with 6.8 & 13.2 I get 0.515 exactly).

    Well, if I'm not mistaken, the truck's speed, relative to the cyclist, should be: VT-C = VT - VC ? I'm not really good with relativity thus far, and I haven't reached the chapter where it goes in-depth, so I'm stuck in a weird place where I can't move to the advanced theory, but I can't understand it with the limited info the book has. Are there any links that simplify relativity in such simple cases?
     
  8. Jul 15, 2017 #7

    TSny

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    We're just considering Galilean relativity here, not Einstein's theory. In many introductory courses, relative velocity is covered as part of the kinematics portion (boats crossing a river with current, etc.)

    But you are right, the relative velocity is VT - VC . So, from the cyclists point of view, it is the same as if she were standing on the ground with the trucks passing her with this relative velocity. You know the distance between the trucks.
     
    Last edited: Jul 15, 2017
  9. Jul 15, 2017 #8
    Okay, so VTC = 15.2 m/s. But how does that help me?

    Can't I just use the 6.8/13.2 explaination as an answer? Is it wrong?
     
  10. Jul 15, 2017 #9

    TSny

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    Imagine you are standing on the ground and the line of trucks is passing you at 15.2 m/s. How often does a truck pass you?

    It's correct. Nothing wrong with it. I don't think too many people would work the problem that way. But that's not a criticism.
     
  11. Jul 15, 2017 #10
    I'm sorry, but I still don't get it. If say, I'm standing at 0 on a horizontal line, and each truck moves at 15.7 m/s, what can I decude from this? I mean, I don't have any other info, do I? Just the speed?

    Oh, okay.
     
  12. Jul 15, 2017 #11

    TSny

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    You know the distance between two consecutive trucks. How much time does it take the trucks to travel this distance if they are moving at 15.2 m/s?
     
  13. Jul 15, 2017 #12
    Uhm, at the risk of sounding like even more of a clueless idiot... do I? Is it the 3546 m I found before? 2 Trucks arrive at the same time, so the other 2 behind them are 3546 meters behind? Is that it?
     
  14. Jul 15, 2017 #13

    TSny

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    Yes, see the picture in post #3. You can see that the distance between two consecutive trucks is half of 3546 m.
     
  15. Jul 15, 2017 #14
    Ah, yeah. But why should it be half of 3546 m? Don't 2 trucks cross 3546 m every 3 minutes? So if the first two arrive consecutively at the dump, the other two will be behind them by 3546 meters. Am I missing something? I figure the "2 trucks at the dump every 3 min" means that the line is compromised ot 2 trucks, side by side, following each other, as in Truck/Truck-Truck/Truck-Truck/Truck. Not Truck-Truck-Truck.
     
  16. Jul 15, 2017 #15

    TSny

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    The problem says, "Garbage Trucks, one after the other, ...". That makes me think one at a time rather than side-by-side. However, it shouldn't make a difference. If you pair them, then the distance between the pairs is 3546 m. Then you use the relative velocity to figure out the time interval between two consecutive pairs passing the cyclist. From that you can get the frequency of pairs. The frequency of trucks would then be twice this frequency. You should get the same answer.
     
  17. Jul 15, 2017 #16
    I figured that it's a case of pairs, otherwise 2 wouldn't reach the dump at the same time. Anyway, so:

    Each pair is seperated by 3546 m. Each pair needs 233 s or 3.88 min to cross them (VTC/d). So, every 3.88 min, 2 trucks overtake the cyclist. So, every 1 min: f = 2/3.88 = 0.515. Which fits with the book's answer and my previous one.

    Thanks a ton for the patience and help!
     
  18. Jul 15, 2017 #17

    TSny

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    OK. Good work.
     
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