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Physics Kinematics Question

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A car is going at 90 km/h in a straight line. Calculate the (uniform) deceleration (in m/s2) needed for it to stop moving at exactly 50 m from the original position. Also calculate how long will it take it to stop (the time it's been decelerating).

    2. Relevant equations

    [tex]x=x_0+v_0t+(1/2)at^2[/tex]

    3. The attempt at a solution

    I'm not exceptionally brilliant at physics or math, so I may have overlooked something which may seem obvious to you, the most experienced ones. Here is my attempt:

    [tex]v_0=90 km/h=25 m/s[/tex]
    [tex]50=0+25t+{at^2}/2[/tex]
    [tex]100=50t+at^2[/tex]
    [tex]at^2+50t-100=0[/tex]

    Now usually I apply here the quadratic formula, but I'm stuck since I do not know the acceleration, nor the time obviously.

    So could someone point to me what did I overlook/miss and what should I do now, please?

    Thank you for your time and patience with me.
     
  2. jcsd
  3. Oct 20, 2011 #2

    Doc Al

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    Staff: Mentor

    You just made a less than optimal choice of kinematic formula. Pick one that has only distance, velocity, and acceleration. (Here's a list: https://www.physicsforums.com/showpost.php?p=905663&postcount=2")

    Alternatively, see what you can deduce from the car's average velocity as it slows down. Use that to find the time and the acceleration.
     
    Last edited by a moderator: Apr 26, 2017
  4. Oct 20, 2011 #3

    Delphi51

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    Homework Helper

    Welcome to PF, Powerof!
    You have an interesting question. The trouble with your solution is that it doesn't include the condition that the car stops at time t.

    You could use V = Vi + at to calculate the time to stop and avoid the whole quadratic calculation. Funny, problems often confuse us with the order of the parts - you have to do the B part before you can do the A.
     
  5. Oct 21, 2011 #4
    Let's see if I understood this well:
    [tex]v=v_0+at \Rightarrow t=\frac{v-v_o}{a}[/tex]
    And now I simply substitute time with it's value ([itex]\frac{Δv}{a}[/itex])?
    [tex]at^2+50t-100=0[/tex]
    [tex]a(\frac{v-v_0}{a})^2+50(\frac{v-v_0}{a})-100=0[/tex]
    [tex]\frac{625}{a}+\frac{-1250}{a}=100 \Rightarrow a=\frac{-625}{100}=-6.25m/s^2[/tex]
    And for the time I guess I apply the formula at the beginning again:
    [tex]t=\frac{v-v_o}{a}=\frac{-25}{-6.25}=4s[/tex]
    As far as I can tell it works:
    [tex](-6.25)4^2+200-100=0[/tex]
    [tex]-100+200-100=0[/tex]
    But now what I really wish to know: was this the thing you intended me to do or did I went for the "less than optimal" again?
     
  6. Oct 21, 2011 #5

    Delphi51

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    Homework Helper

    Looks good. Your solution is more general than I would have done. Using final velocity = 0 early on would reduce the size of the expressions a bit. And it works a little nicer if you go for t first.
    Velocity formula:
    V = Vi + at
    0 = 25 + at
    a = -25/t (1)
    Distance formula:
    100 = 50t + at² (2)
    Sub (1) into (2)
    100 = 50t -25t
    t = 4
     
  7. Oct 21, 2011 #6
    That would indeed be better. Thank you for you help! I really appreciate it.
     
  8. Oct 21, 2011 #7

    Doc Al

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    Staff: Mentor

    I had this formula in mind, which allows you to solve for the acceleration directly:
    [tex]v^2 = v_0^2 + 2 a \Delta x[/tex]

    Alternatively, the average velocity = 25/2 m/s. Which allows you to find the time from the distance (distance = ave velocity * time).
     
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