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Finding the distance, once you have acceleration and time

  1. Apr 11, 2006 #1
    QUESTION: An alcohol free driver needs 0.8 s to stop after seeing an emergency. A person who has had 4 beers need 2 seconds, a person who has had five, need 3 seconds. If driving at speed of 17 m/s, find the reaction distance (after seeing emergency, the distance travelled before pressing brakes) for the all three times. What if the speed is 25 m/s. What if it is 33 m/s.

    so first i had to find acceleration, then the reaction time.

    for the non-alcoholic the acceleration is:
    for 25m/s acceleration = 31.25
    for 33 m/s acceleration = 41.25

    if i have the acceleration how do i find the distance it took to stop. And is this acceleration supposed to be negative?
     
  2. jcsd
  3. Apr 11, 2006 #2

    BobG

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    To find acceleration, I take it you used:

    [tex]v_f = v_i + at[/tex]
    (acceleration is negative)

    To find position, the equation is very similar:

    [tex]s_f = s_i + v_i t + 1/2 a t^2[/tex]

    In fact, the first equation is the derivative with respect to time of the second equation.
     
  4. Apr 11, 2006 #3

    nrqed

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    The question does not ask for the distance to come to a complete stop! It asks "the distance travelled before pressing brakes". There is no acceleration in this problem. They just want the distance travelled before the person starts braking (and during that time, the speed is constant so it's easy!)

    The accelerations you calculated do not make sense. It is as if you are saying "for a alcohol free driver, it wil take 0.8 second to come to rest". This is NOT the meaning of the 0.8 second!

    Patrick
     
  5. Apr 11, 2006 #4

    Hootenanny

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    I hope you mean deceleration :wink:
     
  6. Apr 11, 2006 #5

    arildno

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    Dearly Missed

    Acceleration can perfectly well be negative.

    We use the term deceleration only if we want to emphazise that the SPEED is being reduced.

    When that is not a concern, the term "deceleration" is inappropriate.
     
  7. Apr 11, 2006 #6

    nrqed

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    hear hear!

    The danger is that students make the wrong assumption that deceleration implies negative acceleration and acceleration implies positive acceleration.

    In everyday language, deceleration means that the speed decreases, as pointed out by Arildno. But this may occur for [itex] a_x >0 [/itex] or for [itex] a_x <0 [/itex] depending on the sign of [itex] v_x[/itex]. It is quite dangerous to use the term deceleration with students, in my experience. I prefer to stick with "positive acceleration" or "negative acceleration".

    I like to talk about whether an object speeds up or slows down, but I make it clear to my student that each case may occur with a either positive or negative a_x. The rule being that if both a_x and v_x have the same sign, the object speeds up. If they have opposite signs, the object slows down.

    Pat
     
  8. Apr 11, 2006 #7

    Hootenanny

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    Good point nrged, now I know where all my mistakes come from in kinematics:redface:
     
  9. Apr 12, 2006 #8
    Bravo! So stupid of me to miss something like that :mad: Interestingly worded problem, though :)
     
  10. Apr 12, 2006 #9
    Thank-you so much for your help!!

    Nrged, you are completely right, I misunderstood the question. It is asking for the "reaction distance" so the distance driven before the brakes were applied, which is at a constant speed. So the equation I use is d = v * t .

    Also, how do I draw a vector diagram to scale?
     
  11. May 22, 2010 #10
    QUESTION: An alcohol free driver needs 0.8 s to stop after seeing an emergency. A person who has had 4 beers need 2 seconds, a person who has had five, need 3 seconds. If driving at speed of 17 m/s, find the reaction distance (after seeing emergency, the distance travelled before pressing brakes) for the all three times. What if the speed is 25 m/s. What if it is 33 m/s.

    Cannot be solved: All we know is the initial velocity and how long it took to stop after seeing the emergency. The question asks how long the driver took to react from seeing the accident to pressing the brakes. To solve this we must know the affectiveness of the brakes to determine what part of the stopping time was used by braking and what part was used by getting to the brakes or reacting. The answer will be different depending the affectiveness of the brakes. If when the brakes are pressed it takes .1 sec to stop the car at 17m/s then the reaction time for the alcohol free driver is .7 sec.

    reaction time = total stopping time - braking time.
     
  12. May 22, 2010 #11

    diazona

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    That was my thought as well. I suspect the wording of the question doesn't match the intent of the question, but if it's taken literally as written, I don't see how it can be solved.

    Since the actual stopping time would vary with speed, but there's no mention of which speed the given times apply to, I suspect that those times are reaction times rather than stopping times.
     
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