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Car ride

  1. Sep 24, 2011 #1
    Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 5 m/s2 for 4.3 seconds. It then continues at a constant speed for 9.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 286 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop


    1. How far does the blue car travel before its brakes are applied to slow down?
    2. What is the acceleration of the blue car once the brakes are applied?
    3. What is the total time the blue car is moving?
    4. What is the acceleration of the yellow car?

    1. I tried using 14s to calculate but to no avail.
     
  2. jcsd
  3. Sep 24, 2011 #2

    Doc Al

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    Start by figuring out the motion of the blue car during each segment: acceleration 1, constant speed, acceleration 2. Find the time and distance of each segment.
     
  4. Sep 24, 2011 #3
    i calculate for 14s is the car travel before it slows down is equal to 59.25 is that correct?
     
  5. Sep 24, 2011 #4

    Doc Al

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    No.

    Show how you did the calculation.

    Or just do it step by step:
    How far does the blue car travel during the first 4.3 seconds?
    What is its speed at the end of the first 4.3 seconds?
    How far does it travel during the next 9.7 seconds?
     
  6. Sep 24, 2011 #5
    it travel 1/2*5*4.3=10.25m
    its speed is 5*4.3=21.5m
    for the next 9.7s travel 9.7*5=48.5m
     
  7. Sep 24, 2011 #6

    Doc Al

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    No. The formula should be: x = ½at2.
    Right. 21.5 m/s.
    No. You just said that the speed was 21.5 m/s, so use that fact to calculate the distance.
     
  8. Sep 24, 2011 #7
    ok thank you i got the answer is 254.775m
     
  9. Sep 24, 2011 #8

    Doc Al

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    Good! Keep going.
     
  10. Sep 24, 2011 #9
    2. the acceleration i use d=286-254.775=31.225m
    t=31.225/21.5=1.45s
    then a=21.5/1.45=14.8m/s^2
    is that correct?
     
  11. Sep 24, 2011 #10

    Doc Al

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    OK.
    No, since the speed is not constant. What's the average speed during that segment of the motion?
     
  12. Sep 24, 2011 #11
    ave speed = (0-21.5)/2=-10.75m/s
     
  13. Sep 24, 2011 #12
    i got the answer i use d=31.225,t=31.225/10.75=2.9s
    31.225=1/2at^2
    a=-7.4m/s^2
     
  14. Sep 24, 2011 #13
    acceleration for the yellow car is it related to the blue car?
     
  15. Sep 24, 2011 #14

    Doc Al

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    Good.
    Good.

    You can also use a = Δv/Δt to get the same answer.
     
  16. Sep 24, 2011 #15

    Doc Al

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    Sure. First figure out the answer to part 3.
     
  17. Sep 24, 2011 #16
    total time is 17s
     
  18. Sep 24, 2011 #17
    i can't seem to relate the yellow to the blue car.
    Both have same distance and same time but different acceleration.
     
  19. Sep 24, 2011 #18

    Doc Al

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    Be more accurate than that.

    Use that time and the total distance to figure out the acceleration that the yellow car must have.
     
  20. Sep 24, 2011 #19

    Doc Al

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    Sure, the overall distance and time is the same, but the motions are totally different. The blue car accelerates for a bit, then travels at constant speed for a bit, then accelerates again to slow down. But the yellow car simply accelerates at one rate for the whole distance. And at the end of that time and distance the blue car is at rest but the yellow car is not.
     
  21. Sep 24, 2011 #20
    if i use a=v/t for yellow car,i have to find the v of yellow car then i use the value of v to find a,is that correct?
    But when i get the value for yellow it does not seem right.
     
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