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Constant Acceleration/Velocity HW question

  1. Sep 10, 2008 #1
    I am having trouble with this question. Please help!
    1. The problem statement, all variables and given/known data

    As soon as traffic light turns green, a car speeds up from rest to 50.0 mi/h with constant acceleration of 9.00mi/hr*s. In the adjoining bike lane, a cyclist speeds up from rest to 20.0 mi/h with constant acceleration of 13.0 mi/hr*s. Each vehicle maintains constant velocity after reaching its cruising speed.
    (a) For what time interval is the bicycle ahead of the car?
    (b) By what maximum distance does the bicycle leads the car?

    2. Relevant equations

    xf=xi + vx*t
    vxf= vxi+ax*t
    vx,avg= (vxi +vxf)/2
    xf=xi+1/2(vxi +vxf)*t
    xf=xi+vxi*t+1/2a*t2
    vxf2= vxi2+2*a(xf-xi)

    3. The attempt at a solution

    Im sorry but i have tried every way I can think but I keep on getting the wrong answer. I am completley stuck!
    You dont have to give me the calculations, but just tell me your thought process to get the answers.
    The answer in the back of the book (a) 3.45 s and (b) 10.0 ft

    Please help.
    Thanks!
     
  2. jcsd
  3. Sep 11, 2008 #2

    alphysicist

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    Hi jessedevin,

    Please post your thoughts on this problem and your work that was giving you the wrong answer.
     
  4. Sep 11, 2008 #3
    Well, i know that we have to find where the bicycle and car cross paths, and to do that we have to set there position functions equal to each other. The problem is that I have no clue how to do that. Can someone help me out?
     
  5. Sep 11, 2008 #4
    okay i tried doing it again, but i got the wrong answers. Heres what i did:

    I used the formula xf=xi+vxi*t+1/2a*t^2, because there is a constant acceleration. I used that formula plus the constant velocity formula xf=xi + vx*t for the bicycle numbers and set it equal to the constant velocity + constant acceleration formulas for the car numbers, and solved for t. But I have no clue if its right. Can anyone help?
     
  6. Sep 11, 2008 #5

    LowlyPion

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    Why don't you start out by figuring the things you can?

    How long does it take the bikeguy to get to speed? The car?

    Where will they be when they reach constant velocity? These answers will help you figure out what to calculate next.
     
  7. Sep 11, 2008 #6

    LowlyPion

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    Flesh it out a bit with the numbers. Sometimes just explaining it helps you to understand it better.
     
  8. Sep 11, 2008 #7
    heres what I did:

    postion of car xf =xi+vxi*t+1/2a*t2 t<T
    xf =(xi+vxi*t+1/2a*t2) + (xi + vx*t) t>T

    position of bicycle is the same.

    I am assuming that the bicycle and car intersect aft t>T, so i used the second formula for xf

    Now with numbers

    xf,car = xf,cycle
    0 + (0)t + 1/2(9)t2 + 0 + 50t = 0 + (0)t + 1/2(13)t2 + 0 + (20)t
    t = 15 s, which is way wrong, so i need help.
     
  9. Sep 13, 2008 #8
    No one can help me out? Not even how to start?
     
  10. Sep 13, 2008 #9

    LowlyPion

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    I thought I gave you some pretty good hints. Figure it out a step at a time.

    Let me get you started. How far from the start is he just starting to go at top speed? Would that be v2/2a = 202/2*13 =15.38m from the start? How long to get to speed? V/a = 1.538 secs

    Now you have an equation for the BikeGuy's position as a function of time - after he has gotten to speed - for it must be after he gets to speed since he accelerates faster than the car.

    X = 15.38 m + 20 (t + 1.538)

    Next write an equation for the car's location as a function of time, after determining whether the car is still getting to speed when they pass or whether the car gets to speed before passing. (Figure the car's time to speed and distance to speed as a check.)

    Then you can solve for the case where the car is accelerating, and if the time is greater than the time to get to speed then you must make another equation that takes that into account.

    When their positions are equal - set one equation equal to the other, you will have an equation that you can solve for the Time to Passing.

    Plug that back in to either equation to figure the position.
     
  11. Sep 13, 2008 #10
    Im sorry but Im still confused. Im up to you until we have to write the equation for the position of the car ( I did the previous calculations by myself, and I got the same results). How do you figure out if the car is still getting to speed when they pass or whether the car gets to speed before passing? This is the part the I was confused on since the beginning. I'll show you what I did, but I already know its wrong:

    After finding t=1.54 sec, I found the cars velocity at that time:
    vxfcar= 0+(9)*(1.54)= 13.9 mi/hr
    Then I found the position of the car using the velocity:
    xfcar=xi+1/2(vxi +vxf)*t
    xfcar= 0 +1/2(13.9)(1.54)=10.7 mi
    Then I found the velocity of the car to get to 15.4 mi by using:
    vxfcar2= vxi2+2*a(xf-xi)
    vxfcar2= (13.9)2+ 2(9)(15.4-10.7)
    vxfcar= 16.7 mi/hr
    Then I found the time to get to that velocity by using:
    vxfcar= vxi+ax*t
    16.7= 0 + (9)t
    t=1.85sec
    Then I added the two times and I got ttotal=3.4, but I'm pretty the way I got the answer is not the same way you described. And I do not know how to find the distance the bike lead, even though I know you have to change the units because the book decribes the car in miles but the answer is in feet!
    PLEASE HELP ASAP! I need to figure this out by the end of the weekend!
     
  12. Sep 13, 2008 #11

    LowlyPion

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    Sorry, I didn't notice it was mi/hr. That changes the numbers, but not the method. To keep things simple then I would convert to ft/sec. Using 1.46666 as the easy conversion (5280/3600) and applying that to all the mi/hr I get
    13 m/hr*s = 19.
    9 m/hr*s = 13.2
    20 m/hr = 29.3
    50 m/hr = 73.3
     
  13. Sep 13, 2008 #12
    But Im still confused on your method on getting the time at which the bike leads and the distance during that time. Can you go over it once more. I really need to finish this problem before is due and I think this problem is just over my head.
     
  14. Sep 13, 2008 #13

    LowlyPion

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    Let me do it again. How far from the start is he just starting to go at top speed? Would that be v2/2a = 29.32/2*19. =22.64 f from the start? How long to get to speed? V/a = 29/19 = 1.54 secs

    Now you have an equation for the BikeGuy's position as a function of time - after he has gotten to speed - for it must be after he gets to speed since he accelerates faster than the car.

    X = 22.64 f + 29 (t - 1.54) f
    EDIT: It's (t -1.54) because the first 22.64 is accounted for in that first 1.54

    Next write an equation for the car's location as a function of time

    x = 1/2 a t2 = 1/2 (13.2) t2 = 6.6 t2

    Now set these 2 x equations = to each other.
     
  15. Sep 13, 2008 #14

    LowlyPion

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    Now the time for the car to get to speed is 50/13.2 = 3.78

    So if it looks like it takes longer than that we need a new equation for the car.
     
  16. Sep 13, 2008 #15
    I think you mean 73/13.2 =5.55s...
    but Im think I know where you going with this. But im still working on it
     
  17. Sep 13, 2008 #16

    LowlyPion

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    Absolutely right. I can't lose the old numbers no matter how hard I try it seems. Thanks.

    You're not as lost as you were thinking at first then.
     
  18. Sep 13, 2008 #17
    Okay so I did your method, but I modified some of your numbers to three sig figs...

    So I set the postion formula for the bike equal to the position formula for the car:
    6.60t2= 22.6 + 29.3(t-1.54)
    t= 3. 45 s
    Yes!!! I got that part, but how do I find the maximum distance the bike lead?
     
  19. Sep 13, 2008 #18

    LowlyPion

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    OK almost there then.

    You have two equations of X. What is their difference? When was it at a maximum?

    Happily the difference function is the Quadratic that you just solved. Where they meet is where that function was equal to 0.

    But what you want now is where that function is at maximum. If you have a quadratic where is the maximum?
     
  20. Sep 13, 2008 #19
    I have no clue? What do tou mean by the difference function?
    Do you mean 0=22.6 +29.3(t-1.54) - 6.6t2

    I guess then you find the maximum of this function by taking the derivative and setting it equal to zero to find the time where the velocity is 0. Then put that time into the difference function and that will give me the most distance the bike leads?
     
  21. Sep 14, 2008 #20

    LowlyPion

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    Yes and no.

    I would state it more generally that:
    ΔX = Xb-Xc = 22.6 +29.3(t-1.54) - 6.6t2

    The question asked you to solve for when ΔX = 0. Which you did.

    But now you want the maximum of ΔX and as you said you take the derivative ... That part you know what to do. I just wanted you to clearly understand why you can do that is all.

    Good job.
     
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