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Stopping distance

  1. Feb 18, 2009 #1
    Hi,

    There is a cyclist rolling but not peddelling, so friction will eventually cause him to stop. You dont know a start speed, but Every 10 metres a time reading is taken to see how long it took him to travel that last ten metres so an acceleraton can be calculated

    relative to when readings were taken:
    0 - 10m : 2.5seconds
    10 - 20m : 3.5 seconds

    How do I work out how many metres he will travel?

    Can I do a = [(distance/time1)-(distance/time2)]/total time

    i.e [10/2.5 - 10/3.5]/(2.5 + 3.5)

    = acceleration of -0.19 metres/second^2

    Is this correct???
     
  2. jcsd
  3. Feb 18, 2009 #2

    rcgldr

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    Assuming a constant rate of decleration, then velocity change per unit time is linear.

    Those ratios you showed are the average velocity during the two stated intervals.

    10 m / 2.5 s = 4.0 m/s

    10 m / 3.5 s = 2.86 m/s

    How long did it take for this change in average velocity to occur?
    (Note it's not the total time).

    See if you can figure out from here.
     
  4. Feb 19, 2009 #3
    Its quite hard to think about - very easy to make yourself believe the wrong thing when thinking about this.

    So, because they are average speeds, that is like saying that is the speed is 5 and 15m into the 20 metre travel.

    he is covering 4 m/s 5 metres into the 20m, and 2.86m/s 15m into the 20 metres.

    So if i took the average of these two speeds, that is the average number of metres he covers a second, between those two points (between the 5 an 15metre marks),

    so his average velocity over the whole 20metres is 3.43 m/s

    ??????

    But if this is correct how can i think of it as an acceleration so that i can work out his speed in 50m, or I can work out how far he will travel in total after the first stopwatch is started?

    I'm struggling to visualze what could be going on, sorry
     
  5. Feb 19, 2009 #4

    LowlyPion

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    If you are having difficulty with the problem then work it out the long way. For instance you know that for a negative acceleration of slowing

    X = Xo + Vo*t - 1/2*a*t2

    For the first interval since X - Xo is given as 10

    10 = Vo1*2.5 - 1/2*a*(2.5)2

    For the second interval you know that its Vo2 is the Vo1 of the first -a*t as it slowed over the first 10m. So for the next interval ...

    10 = Vo2*t - 1/2*a*t2

    which becomes

    10 = (Vo1 - 2.5*a)*(3.5) - 1/2*a*(3.5)2

    Happily now you have 2 equations and 2 unknowns.
     
  6. Feb 19, 2009 #5

    rcgldr

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    I'm not sure about the wording of the problem here. If you're taking readings from a stop watch is the stop watch reset at each interval or allowed to continue running, and starting from a non-zero value? If this is the case, then the unknown reading at 0 m is t0 the time it takes to go 0 to 10m t1 and the time it takes to go 10m to 20m t2.

    0 m -> 10 m : reading = t0 + t1 = 2.5 seconds
    10 m -> 20 m: reading = t0 + t1 + t2 = 3.5 seconds
    and t2 = 1 second

    I'll skip the solution for this case for now.

    I'll assume the simple case that the stop watch is reset every 10 m then you have
    t1 = 2.5 seconds
    t2 = 3.5 seconds

    Since the decleration is linear, the velocity at the middle of the time interval = the average velocity.
    v1 = (10 / 2.5) (m/s) = 4.00 m/s at time 1.25 (0.0 + (2.5 / 2))
    v2 = (10 / 3.5) (m/s) = 2.85 m/s at time 4.25 (2.5 + (3.5 / 2))

    The change in velocity (4.00 m/s - 2.85 m/s) occured in 3.00 (4.25-1.25) seconds. What does that tell you about acceleration?
     
    Last edited: Feb 19, 2009
  7. Feb 19, 2009 #6
    Yeah sorry I meant that the time for each ten metres HAD started again at zero on the stopclock.

    The acceleration is (-1.15/3) m/s/s

    or 0.383m/s/s

    And this is linear. In other words I can say that the last average time recorded was 2.85m/s and the cyclist is slowing at 0.383m/s/s so

    constant decceleration, therefore s=uv-0.5at*t

    4*2.85 - 0.5*0.383*3*3

    therefore the cyclist will travel another 9.68m???

    Am I getting there?

    If I then said the cyclist was steering in circles, I could divide this by 2PI, and see the number of radians left in his travel before he stops?
     
  8. Feb 19, 2009 #7

    LowlyPion

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    You must be on the right track because .38 agrees with the solution made the long way I suggested.

    To figure the distance just figure the Vo at the first measurement interval that comes readily from plugging into the first equation namely that

    Vo = 10 + 1.25*a = 4 + 1.25(.38) = 4.475 m/s

    From there then it's easy to figure the total distance as

    (4.475)2 = 2*a*x

    where x is the distance from the beginning of the first interval.

    That works out for me as 26.34 less the 20m of the 2 intervals or 6.34 m.
     
    Last edited: Feb 19, 2009
  9. Feb 19, 2009 #8
    Ok, I'll do this again on paper, tomorrow, but am I right, in thinking
    "If I then said the cyclist was steering in circles, I could divide this by 2PI, and see the number of radians left in his travel before he stops?"

    ???
     
  10. Feb 19, 2009 #9

    LowlyPion

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    The distance / 2πr is the number of revolutions of the wheel, if that's what you're getting at. But you will need r to determine revolutions or radians.
     
  11. Feb 19, 2009 #10

    rcgldr

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    Since the decleration is linear, the velocity at the middle of the time interval = the average velocity.
    v1 = (10 / 2.5) (m/s) = 4.000 m/s at time 1.25 (0.0 + (2.5 / 2))
    v2 = (10 / 3.5) (m/s) = 2.857 m/s at time 4.25 (2.5 + (3.5 / 2))

    The change in velocity (4.000 m/s - 2.857 m/s) occured in 3.00 (4.25-1.25) seconds.
    acceleration = -1.143 m / (3 s2 ) = - .381 m / s2

    Where did you get the 3 seconds of time from? How much distance was covered as the bicycle slowed from 4.00 m /s to 2.85 m / s?

    acceleration = - .381 m / s2

    average velocities
    va1 = (10 / 2.5) (m/s) = 4.000 m/s at time 1.25 (0.0 + (2.5 / 2))
    va2 = (10 / 3.5) (m/s) = 2.857 m/s at time 4.25 (2.5 + (3.5 / 2))

    velocity at the end of last interval:
    ve2 = 2.857 m/s + ( - .381 m / s2 ) x (+ 3.5 s / 2) = 2.190 m / s
    How much time does it take to stop from 2.190 m / s and how much distance is covered in this time?

    or velocity at the beginning of first interval:
    vb1 = 4.00 m/s + ( - .381 m / s2 ) x (- 2.5 s / 2) = 4.476m / s
    How much time does it take to stop from 4.476 m / s and how much distance is covered in this time?
    How much distance past 20 m is coverered in this time?
     
    Last edited: Feb 20, 2009
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