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Homework Help: Kinematics Car Distance Problem

  1. Jan 31, 2006 #1
    I have a problem that I just cant get a handle on.
    You have a car that drives from A to B in 1620 seconds and then B to C.
    The total distance from A to C is 10,000 meters (A to C is a straight line)
    The acceleration of the car from B to C is .2 m/s^2
    I need to know how long the car takes to go from B to C IF it it travel from
    A to C in a total time of LESS than 1800 seconds (30 minutes)
    the distance from A to B is 8900 meters and B to C is 1100 meteres

    Thank you

    blumfeld0
     
  2. jcsd
  3. Feb 1, 2006 #2

    tony873004

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    IF it it travels from
    This is like saying the car drives from A to C doing at least 20 km/hr

    So your question becomes:
    How long does it take a car to travel 1100 meters given that its initial velocity is at least 20 km/hr and it is accelerating at 0.2 m/s^2?

    I'd still need to know what initial velocity is. Without that, I could only give you a range of answers, using 20 and infinity as velocity to compute the answer.

    Maybe your answer is supposed to be in the form:
    "The car will take no longer than ____ to travel from B to C."
     
    Last edited: Feb 1, 2006
  4. Feb 3, 2006 #3
    well, I might be wrong but i would say it would take about 50.74 seconds. First, i found the velocity of the car from a to b (v=d/t), knowing that i had the initial velocity to start thinking about the problem from point b to c. Then i use the formula vf^2=vi^2+2ad for and I plugged in for vi= 5.49 m/s a=.2m/s^2 (given) d= 1100m (from point b to c) then i would have the final velocity with that i used v=d/t and changed it to t=d/v and i got 50.74 seconds. I might be wrong and if you find the right answer let me know
     
  5. Feb 3, 2006 #4

    tony873004

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    I think you got it wrong. Disregard my original post. I fell for the irrelevant data trick.

    Once you find your final velocity, compute your average velocity (B to C) and use that in your t=d/v formula.

    Or use the formula xf=xi+vit+0.5at^2 and use the quadratic equation. I get the same answer both ways.
     
  6. Feb 4, 2006 #5
    It is ok (in a way) if I got it wrong. It is all about knowing how to do it. However, what irrelevant data are you talking about?..... I got vf=21.68, i know that you said that after you plugged the final velocity into both of the formulas you got the same answer, however, i did not, so yes, i see that i could have made a mistake. It would be helpful if you would tell me perhaps how you achieved your answer and ofcourse state all of the calculations you made, that way i could compare, understand, and learn how to do the problem. Thank you.
     
  7. Feb 4, 2006 #6

    tony873004

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    What answers did you get?

    The irrelevant data is this: "it travel from A to C in a total time of LESS than 1800 seconds".

    I didn't use 1800 for anything.

    No, I didn't plug the final velocity into either formula. The 1st formula (t=d/v) wants average velocity for v, and the 2nd formula wants initial velocity.
    Average velocity = (final velocity + initial velocity) / 2

    Did you try the quadratic way?
     
  8. Feb 4, 2006 #7
    Everytime i use a different formula i get a different answer which has confused me in a way, quadratic way? if by that you mean "xf=xi+vit+0.5at^2" what is x? distance? if it is then i have never seen such equation before and it gives me a negative distance which i cannot take the sq root of since it would be a non real answer. For such reason, i dont believe that x would stand for distance. After using some other formulas i have gotten 81 seconds using avg velocity. I do not understand why you did not use avg velocity.
     
  9. Feb 4, 2006 #8

    tony873004

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    I got 81 too.

    I can't seem to get TEX to work in preview mode, so this will be ugle

    The quadratic way: (this way is harder, but it's good to know how)

    xf=xi+vit+0.5at^2

    1100=0+5.49t+0.5*0.2t^2

    set it equal to 0

    0.1t^2+5.49t-1100=0

    a=0.1
    b=5.49
    c=-1100

    t=(-b+-sqrt(b^2-4ac))/2a

    Since it's + or - sqrt, you need to do it twice and discard the unreasonable answer:

    t=(-5.49+sqr(5.49^2-4*0.1*-1100))/(2*0.1)

    t=80.96 (accept this answer since it agrees with your other answer and it seems reasonable)

    or

    t=(-5.49-sqr(5.49^2-4*0.1*-1100))/(2*0.1)
    t=-135.9 (disragerd this answer since your time can't be negative)
     
  10. Feb 4, 2006 #9
    Thank you very much, tony873004, for i have learned some things thanks to you. Things that i had not learned or perhaps that i had not paid enough attention to, which i regret.
     
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