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Motion in a straight line tough one

  1. Oct 6, 2010 #1
    motion in a straight line....tough one

    ok, a cyclist is travelling at 10ms^-1. he sees trafic lights ahead at 108m. He carries on cycling for a time T before decelerating and stopping at the lights 16.7seconds after he first saw the lights.

    Calculate T

    Im a bit stuck because i feel this question doesnt give enough information
    If anyone could point me in the right way it would be much appreciated :)
     
  2. jcsd
  3. Oct 6, 2010 #2

    berkeman

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    Staff: Mentor

    Re: motion in a straight line....tough one

    I agree that it seems that you need more information to calculate T. The rate of deceleration would impact how long the constant velocity could be kept up.
     
  4. Oct 6, 2010 #3
    Re: motion in a straight line....tough one

    The question actually says
    2 cyclists (B) travelling at 9ms^-1 and (A) travelling at 10ms^-1 are at the same point when they see traffic lights 108m away. B travels at its speed for a furthe 6seconds before decelerating and stopping at the lights. A travels for a time T before decelerating and reaching the lights at the same time as B

    Calculate T

    I worked out it took B 16.7 seconds to stop so figured that it has to be the same for A

    Thats as much as i got :(
     
  5. Oct 6, 2010 #4

    berkeman

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    Staff: Mentor

    Re: motion in a straight line....tough one

    Could you post your work for B that gave you 16.7 seconds to stop?
     
  6. Oct 6, 2010 #5
    Re: motion in a straight line....tough one

    Hm, maybe you have to assume they can only decelerate at a fixed number lambda, so that you could use the same number from B for A. I did get 12 seconds for B's stopping time though...

    Anyway, do you have access to an answer to know you're right or not?
     
  7. Oct 6, 2010 #6
    Re: motion in a straight line....tough one

    6s*9ms^-1 = 54m
    108m-54m = 54m

    v^2=u^2+2as

    0^2=9ms^-1 + 2a*54

    a = -0.84359.....ms^-2

    t=v-u/a

    t=0ms^-1/0.8435...ms^-2
    t = 10.7s

    total time = 6 + 10.7 = 16.7s
     
  8. Oct 6, 2010 #7
    Re: motion in a straight line....tough one

    Hi how did you get 12?

    no access to answers :(
     
  9. Oct 6, 2010 #8
    Re: motion in a straight line....tough one

    sorry yer got 12 now, me calculating wrong lol

    couldnt i use T and work iit out algebraically in an equation?
     
  10. Oct 6, 2010 #9
    Re: motion in a straight line....tough one

    First of all you forgot the ^2 for the 9m/s, but that might be a typo. Still, if you have -81 = 108a, then a = 0.75

    EDIT: woops, apparently I hadn't seen the last post before posting this...

    And what do you mean "use T"? How can you use what you want to find? Anyway, if you can't assume A and B experience the same value for their acceleration, I think you simply don't have enough information to solve for T...
     
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