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Homework Help: Beginner Physics Kinematics/Motion do not have a physics brain

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data
    a motorist is driving at 20m/s when she sees that a traffic light 200m ahead had just turned red. she knows that this light stays red for 15s, and she wants to reach the light just as it turns green again. it takes her 1.0s to stop on the brakes and begin slowing. what is her speed as she reaches the light at the instant it turns green?
    The answer is 5.71 m/s.

    2. Relevant equations
    d= ( v1+v2/2) (t)

    d= v1(t)+1/2(a)(t2)

    3. The attempt at a solution

    Tried to find v2 by rearranging the first equation.. tried finding a to possibly use it as another variable in an equation.. nothing is working
  2. jcsd
  3. Jan 11, 2009 #2
    I can't follow your question.
    you mean you want the acceleration if she(he) can reach the light at the instant it turns green?
  4. Jan 11, 2009 #3
    no i want the velocity
  5. Jan 11, 2009 #4
    I believe your missing a variable here...
  6. Jan 11, 2009 #5
    so im not completely hopeless?
  7. Jan 11, 2009 #6
    First of all, I don't think there is such a thing as a physics brain. lol It's just a lot of practice and over time, you get a hang of things.

    Well, I did get a solution, but not 5.71. Where did you get the answer from?

    vo=20 m/s
    x=200 m
    t = 15 sec (but since it takes 1 second to react, 14 sec is the actual time)

    so I used
    x=vot+ 0.5 at2
    200 = 20(14) + .5 a (14)2
    solve for a

    then use Vf=Vo+at

    But I got 8.xx for the final velocity. Not sure if I am interpreting all the values right. Maybe...
  8. Jan 11, 2009 #7
    yeah,it doesnt make sense.maybe you misunderstood your question.read it again!
  9. Jan 11, 2009 #8
    well look at your equation. I find that if you do something like this it helps to work out your answer.

    V1= ?
    V2= ?
    a = ?
    t = ?
    d = ?

    This gives you quick reference at any time to any of your variables and allows you to do conversions beside as needed.

    so as an example if say I was to use the equation:

    V2^2 = V1*t +0.5 * a * t^2

    I would do say

    V1= 2
    V2= ?
    a = 3
    t = 5
    d = n/a

    also remember for any linear motion question like this you need 3 variables given.
  10. Jan 11, 2009 #9
    oh trust me, no amount of practise will ever make me better at this

    and I dunno .. i have a test tomorrow and i wanted extra problems and someone got one from someone and thats what they sent me..
  11. Jan 11, 2009 #10
    v1 is 20 m/s
    where did you get 3 for a?
  12. Jan 11, 2009 #11
    Well remember that if you can't do pure physics Engineering is always an option...
  13. Jan 11, 2009 #12
    v1= 20 m/s
    d=200 m
    t= 14 s
    a= ?
    v2= ?
  14. Jan 11, 2009 #13
    those were random numbers I'm not doing the work because I'm doing engineering equations right now sorry if I confused you...
  15. Jan 11, 2009 #14
    Think of it this way, your primary goal here is to have her end up 200m down the road in 15 seconds (also take into consideration that it takes her 1 second to start to slow down), which would put her right at the light when it turns green. So you know that her Vi=20m/s.

    So start this way, see how long it takes her to get 200m down the road going at a constant 20m/s. If its under 15s, she is going too fast, if it is over 15s, she is going too slow.

    So once you see that, how many seconds is she off by? Then take acceleration (or deceleration) into account and see what constant deceleration would put her exactly at 15s. Then you can find her final Vf at 15s.
  16. Jan 11, 2009 #15
    Hahaha, well. Personally, it took me about 2 weeks to master this stuff. Then rest of physics came easier as you progress.

    Don't worry, you will get a hang of it. As long as you realize that physics isn't about the equations. The mathematics are just tools, the true essence is knowing how to apply the tools.

    Edited the original post with my solution...which isn't really the same as your answer. >.<
  17. Jan 11, 2009 #16
    i make it
  18. Jan 11, 2009 #17
    but answer is 6.28m/s
  19. Jan 11, 2009 #18
    Did you take t = 14 or 15?

    Because if you take 1 second to step on the break...then the t should be 14.
  20. Jan 11, 2009 #19
    completely irrelevant but anyone else think 1 second is a slow reaction time?
  21. Jan 11, 2009 #20
    yes,after 1 second:
  22. Jan 11, 2009 #21
    Oh nice. I forgot about the distance difference. ^.^ Yep. Hehehe, my mistake there.
  23. Jan 11, 2009 #22
    Maybe there are mistakes in my solution too.
    I also make mistakes like this again and again. I just cant help it and have been accustomed to it.
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