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Car stopping time and distance when speed is doubled

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A. A car moving at speed v can stop in distance d in time t. The same car moving on the same road at twice the speed takes what distance and what time to stop?
    Answer: a) Distance d and time t/2. b) Distance 2d and time t. c) Distance 4d and time 2t. d) Distance 2d and time 2t. e) Not possible to determine.

    2. Relevant equations
    A. Vf = Vi+2ad?
    B. W = F*D?

    3. The attempt at a solution
    A.I assume that you would have to use velocity equation to figure this out, but I don't understand how you can use the equation if there is no numbers given....
    B. I tried to find the work of the obj then set it equal to the work done on the person (pulling) and devide by force to find distance, and got 20.... is this not the right equation?
     
    Last edited by a moderator: Nov 23, 2016
  2. jcsd
  3. Nov 23, 2016 #2
    Problem A:
    It should be: vf2 = vi2 + 2ad
    That is a good equation to use for distance.

    For these kinds of problems, you just have to start by writing out the equations using the variables (no values) for the two different cases - case 1 and case 2 - using appropriate subscripts for case 1 and 2. You also have some additional information that the initial velocity in case 2 is twice that of case 1, which you can put into a simple equation and substitute into one of the other two equations. So what is the key variable that you are interested in? In this case it is d, so you want to solve both equations for d. Then in order to find the ratio of the two distances, you can divide one equation by the other. So you should end up with d2/d1 equal to some numeric value. That's how you have to solve these kinds of ratio problems. Then you will have to use a different equation to find the ratios of the times t1 and t2.
     
  4. Nov 23, 2016 #3
    so basically I plug in what info I have for each system, then solve for d and divide each equation by it's self?
     
  5. Nov 24, 2016 #4
    There is no need for numbers
    use the simple equation d=vt
    now duplicate v so that V=2v.............V, D and T will be the new velocity, distance and time
    now try to substitute in every choice V and the time given
    a) D=V.T D=2v*t/2=v/t=d ......................... so d and t/2 is correct
    and continue in that way
     
  6. Nov 24, 2016 #5
    d = vt applies for a constant velocity (acceleration = 0). This situation is a car that is decelerating - braking from some initial velocity to a stop (v=0).

    I wouldn't say "divide each equation by itself"; I would say "divide one equation by the other".
     
  7. Nov 24, 2016 #6

    PeroK

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    It is important to learn to work with algebra not just numbers. However, let me show you one thing that helps in these problems.

    Just pick any speed for ##v## and any braking deceleration ##a## and calculate ##d## and ##t##.

    You might as well pick some easy numbers, so try ##v = 20m/s## and ##a= -10m/s^2##.

    Then repeat the calculation using twice the speed ##v = 40m/s## and calculate ##d## and ##t## for this case.

    That will actually give you the answer - you just need to compare the answers in the two cases. Or, you could use this method to check an answer you got using algebra.
     
  8. Nov 24, 2016 #7
    how can you tell it decelerating?
     
  9. Nov 24, 2016 #8
    Because of the following wording in the problem statement:
    The car goes from some initial velocity, v, to being stopped (v=0).
     
  10. Nov 24, 2016 #9
    but you dont have to take that into consideration if it doesnt mention that do you?
     
  11. Nov 24, 2016 #10
    I'm sorry, but I don't understand the point you are making in your statement above. If the problem doesn't mention that the car goes from some initial velocity to velocity = 0, then that would certainly have to be considered. But if that point was removed, it would be a different kind of problem altogether and more information would be needed.

    This problem is an exercise in determining stopping distance relative to initial velocity. A car is moving along at some initial velocity and the driver slams on the brakes. How much time is required for it to stop, and what distance is required for it to stop. And how does that time and distance compare to that same car if its initial velocity is doubled? One might be inclined to think, "If the speed is doubled, the stopping distance would also be doubled and the time required to stop would be doubled." So if one works out the equations, that assumption will be proven or disproven.
     
  12. Nov 24, 2016 #11
    what I mean is that for example, if a problem didnt mention about friction, then you dont need to calculate for friction too, but I see what you mean. also I tried to use the original equation posted and set the the 2x velocity to be 2v an device an equation by the other and I get either 1/2 or 2 since the variables cancels out...
     
  13. Nov 24, 2016 #12
    You get 1/2 or 2 for what parameter - distance or time?
     
  14. Nov 24, 2016 #13
    either one, unless Im doing it wrong.
    vf2 = vi2+2ad
    ----------------------------------------------------------- (divide)
    vf2 = 2vi2+2ad

    the velocity, accel, and distance cancel out (doesnt really matter which variable I bring to the left side since It'll end up being the same). I think Im translating what they mean by diving the equation by the other wrong...
     
  15. Nov 24, 2016 #14
    It should be vf2 = (2vi)2 + 2ad
     
  16. Nov 24, 2016 #15
    Ok, Im still stump.... I don't know what Im doing wrong. but I'm ending up with 1/4 (same as before where the var cancels out... unless Im doing my algebra wrong........
     
  17. Nov 24, 2016 #16
    I think 1/4 (or 4) is the right answer for distance. As the velocity doubles, the distance required to stop increases by a factor of 4.
    So what about time? How much does it increase as velocity is doubled?
     
  18. Nov 24, 2016 #17
    using vf=vi+2at, I got 2 or 1/2 (idk which divide by which)
     
  19. Nov 24, 2016 #18
    Yes, 2 looks like the right answer for time. It's pretty easy to figure out which time is longer. Also, it's a good idea to use subscripts so you can keep track of things. So you might write:

    v1f = v1i + 2at1
    v2f = v2i + 2at2

    Of course, the final velocity will be 0 in both cases.
    And for case 2, the initial velocity is twice that of case 1, so v2i = 2v1i
    Then when you divide that out you should get t2/t1 = 2
     
  20. Nov 24, 2016 #19
    Ah sorry I didn't pay attention for that
     
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