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Guys, I have 4 easy questions. (probably easy to you)

  1. Nov 7, 2007 #1
    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.
     
    Last edited: Nov 7, 2007
  2. jcsd
  3. Nov 7, 2007 #2
    Well just use newtons equations of motion, which are:

    v = u+at

    v^2 = u^2 + 2as

    s = ut + 0.5a(t^2)

    where u = initial speed, v = final speed, s = distance, t = time, a = acceleration

    Then just need to look at each situation, break it down, and apply soem of the equations.
     
  4. Nov 7, 2007 #3
    Thats my main problem. I dont know which equation to use.
     
  5. Nov 7, 2007 #4
    Make a list of the variables you have and what you're trying to solve for, and find the equation that applies.

    Take 3a) We know the time (t), the initial velocity (vo), and the final velocity (v). We're looking for acceleration. Which equation would you use?
     
  6. Nov 7, 2007 #5
    For #1, should I be using
    d = Vo t + ½ at2

    ??
     
  7. Nov 7, 2007 #6
    To find the distance of the surface of the water to the diver when he releases the cork, yes.
     
    Last edited: Nov 7, 2007
  8. Nov 7, 2007 #7
    For example the first answer is as follows:

    1. Find depth of the diver when cork is released

    so we work out distance travelled by the cork with

    a = 3
    t = 2
    v = ?
    u = 0
    s = ?

    we want s so pick equation that has s as the only unknown in it

    s = ut + 0.5a(t^2)

    so

    s = 0.5(3)((2)^2)
    s = 6 m

    Now find distance travelled by diver in the 2 seconds

    s = ut as we know a=0
    s = (1)(2) = 2m

    So total depth is 6+2 = 8m
     
  9. Nov 7, 2007 #8
    why would u be 0? Isnt the diver going at 1.0m/s
     
  10. Nov 7, 2007 #9
    Alright, I solved it in my head but I just used my common sense tosolve the question. Simple mental math.

    I did get 8. I just hate these formulas

    idk about the other 3 tho
     
  11. Nov 7, 2007 #10
    It's the starting velocity of the cork, not the diver. By solving the equation you're getting the distance to the diver at the depth where he releases the cork. You also need to take into account the distance he moves in the time it takes the cork to go up, that's where we get the extra 2m.
     
  12. Nov 7, 2007 #11
    I figured it out. Now what about the next 3? I feelso ignorant now.
     
  13. Nov 7, 2007 #12
    Actually i made mistake :P

    The cork does have an initla velcocity of 1m/s in the downwards direction as it is moving in the same direction as the diver. So first it must deccelerate before it can move in the surface direction. Taking upwards to be the positive direction we must use :

    v = ?
    u = -1 negative as it is in downwards direction
    a = 3 positive as it is in upwards direction
    t = 2
    s = ?

    s = ut + 0.5a(t^2)
    s = (-1)(2) + 0.5(3)(2^2)
    and so s = 4m

    and so we get 2+4 = 6m depth
     
  14. Nov 7, 2007 #13
    Yea, but when you are descending into the water, you are going down. The diver would be 8m into the water.

    edit-I am not sure myself now. You've got me thinking.
     
  15. Nov 7, 2007 #14

    Astronuc

    User Avatar

    Staff: Mentor

    With respect to problem 1, it would probably be safe to assume the cork is release with zero initial velocity, since the deceleration in water would be very rapid (i.e. much faster than the 3 m/s2. The diver could release it in such away to have a zero initial velocity.

    The the problem is one in which the cork simply accelerates at a constant rate vertically (to the surface) for 2 sec, and the diver continues to descend at constant velocity.
     
  16. Nov 7, 2007 #15
    so you're saying the answer is 8? What are your thoughts about the other problems
     
  17. Nov 7, 2007 #16

    Astronuc

    User Avatar

    Staff: Mentor

    Use V2 = Vo2 + 2ad for problem 2.

    One is given Vo and a. Find the distance in part a.

    Then knowing the distance d and a, one can find Vo.
     
  18. Nov 7, 2007 #17
    I am confused. I dont know which goes where
     
  19. Nov 7, 2007 #18
    Well V is your final speed which is going to be 0, then the Vo is the speed of the car before the braking occurs
     
  20. Nov 7, 2007 #19
    For d on the dry road

    I got 452m

    I doubt that is right..
     
  21. Nov 7, 2007 #20
    Well you already know d for the dry road, what you need is d for the wet road, so :

    V = 0
    Vo = 27 m/s
    a = -6.5 m/s2
    t = ?
    d = ?

    so using V = Vo^2 + 2ad

    -(Vo^2)/2a = d
    -(27^2)/2(-6.5) = d = 56m
     
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