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Questions on Kinematics

  1. Nov 7, 2007 #1
    Im making a new thread because I coulnt edit my previous thread title


    1. The problem statement, all variables and given/known data
    1. While descending at a constant speed of 1.0 m/s, a scuba diver releases a cork, which
    accelerates upward at 3.0 m/s2. What is the diver’s depth when the cork reaches the surface 2.0 s later?

    2. A car with a velocity of 27 m/s slows down at a rate of - 8.5 m/s2 to a stop in a distance
    of 43 m on a dry road. The same car traveling at 27 m/s slows down at a rate of -6.5 m/s2 to a stop on a wet road.
    a. How much farther does the car travel on the wet road before coming to a stop?
    b. What maximum car speed will allow the car traveling on the wet road to stop in a distance of 43 m?

    3. A car traveling at 14 m/s encounters a patch of ice and takes 5.0 s to stop.
    a. What is the car’s acceleration?
    b. How far does it travel before stopping?

    4. An accelerating lab cart passes through two photo gate timers 3.0 m apart in 4.2 s. The velocity of the cart at the second timer is 1.2 m/s.
    a. What is the cart’s velocity at the first gate?
    b. What is the acceleration?





    2. Relevant equations
    WE NEED TO USE ONE OF THESE EQUATIONS FOR THESE QUESTIONS
    V= Vo + at
    d = Vo t + ½ at2
    V2 = Vo2 +2ad
    d = ½ (V + Vo) t

    g= -9.8 m/s2



    3. The attempt at a solution

    I am sorry. I really have no idea. I have been sick for quite some time and missed a lot of school time. Since we are on break, this was our assignment, I cannot ask the teacher for help.

    If you can tell me which equation I need to use, I am sure that will help.
     
  2. jcsd
  3. Nov 7, 2007 #2
    A good way to start any kinematic problem is to write down all the given quantities, and then use an equation that has all the known quantities except one.

    Let's use #3 as an example:

    we know [tex]\Delta[/tex]t = 5 seconds, [tex]V_{1}[/tex] = 14m/s, and since the car needs to stop, it's [tex]V_{2}[/tex] would 0
     
    Last edited: Nov 7, 2007
  4. Nov 7, 2007 #3
    I solved 1-3; I am only in need of help on number 4.
     
  5. Nov 7, 2007 #4
    Luckily I had a question very similiar to that on my test. Here it goes

    In this case, we only need to worry about what happened between the 2 gate timers, everything before/after we can ignore.

    So we know the distance it travelled, d = 3.0m, we know how long it took it to travell 3.0 m, t = 4.2 seconds, and we know it's velocity at the second timer [tex]V_{2}[/tex] = 1.2 m/s. Now we need to find an equation to relate these 3 known variables.

    Another way you can do this, is do part b) first, and then a)
     
  6. Nov 7, 2007 #5
    what did u get to find the accleration
     
  7. Nov 7, 2007 #6
    My mistake, I think it would be better if you did part a) first and then b). Once you do that, post your work here.
     
  8. Nov 7, 2007 #7
    I have no idea what to do..

    I cant even properly set up the equation
     
  9. Nov 7, 2007 #8
    which of these 4 equations has 3 of the known quantities, and 1 unknown quantity that you need?
     
  10. Nov 7, 2007 #9
    the 4th( i think) but when I set it up.. things didnt work out too well
     
  11. Nov 7, 2007 #10
    I think the 4th is right. Let me give a try (hope I'm not wrong)

    d = ½*(V + Vo)*t

    3 = ½*(1.2 + Vo)*4.2
    6 = (1.2 + Vo)*4.2
    1.2 + Vo = 6 / 4.2
    Vo ~ 1.43 m/s

    yea... I see what you mean. It's initial velocity is supposed to be smaller than it's velocity at the second gate...
     
    Last edited: Nov 7, 2007
  12. Nov 7, 2007 #11
    so it is w4ong?
     
  13. Nov 7, 2007 #12
    I'm an idiot

    3 = ½*(1.2 + Vo)*4.2
    6 = (1.2 + Vo)*4.2
    1.2 + Vo = 6 / 4.2
    Vo = 1.43 - 1.2 = 0.23 m/s

    that sounds more possible because the cart is accelarating so it's supposed to gain speed.
     
  14. Nov 7, 2007 #13
    i'll try that

    im not following your work

    I m lost in it
     
    Last edited: Nov 7, 2007
  15. Nov 7, 2007 #14
    where did u get 6 from?
     
  16. Nov 7, 2007 #15
    3 = ½*(1.2 + Vo)*4.2
    6 = (1.2 + Vo)*4.2

    I needed to get rid of the ½ and so I multiplyed both sides of the equation by 2
     
  17. Nov 7, 2007 #16
    why would u do that....
     
  18. Nov 7, 2007 #17
    to isolate Vo

    Here's an example:

    you have the following equation

    ½X = 3

    Isolate X so we would multiply both sides by 2 to isolate X so we get X = 6

    2 * ½ = 1
     
  19. Nov 7, 2007 #18
    i got .48m/s
     
  20. Nov 7, 2007 #19
    it would help if you posted your work
     
  21. Nov 7, 2007 #20
    [​IMG]

    I used the bottom right equation

    and took it from there

    3= 1.2/2 (4.2)

    3-2.52=.48
     
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