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anonymous

  • 4 years ago

A car slows down from 20 m/s to rest in a 85m distance. What was its acceleration if assumed constant?

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  1. anonymous
    • 4 years ago
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    So far I've set up the problem as such: the initial velocity is 20 m/s, the final velocity is 0m/s, the distance is 85m and I want the acceleration.

  2. JamesJ
    • 4 years ago
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    By the Work-Energy theorem, the change in kinetic energy must be equal to the work done: \[ \Delta KE = Work \] Now \( Work = Fd = mad \). You should now have enough to solve for acceleration \( a \).

  3. anonymous
    • 4 years ago
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    I haven't learned that yet. This is using the kinematic equations. I was thinking of using either vf^2=vi^2+2ad where vf is final velocity, vi is initail velocity, a is acceleration and d is distance or d=vi + vf/2 but when I plugged in my values for weach equation, I got 570 and 8.5, which were wrong answers.

  4. JamesJ
    • 4 years ago
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    This is where that equation comes from. \[ \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \] Now by the Work-Energy theorem, the change in energy of a system is equal to the work done on it, \[ \Delta KE = Fs = mas \] where \( s \) is displacement. Hence we have that \[ \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mas \] Cancelling now \( m \) from both sides and multiplying through by 2, we have \[ v_f^2 - v_i^2 = 2as \]

  5. JamesJ
    • 4 years ago
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    Or in other words, \[ v_f^2 = v_i^2 + 2ad \] writing now \( d \) for displacement. In your problem, you know everything in this equation except acceleration \( a \).

  6. JamesJ
    • 4 years ago
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    Because \( v_f = 0 \), you have \[ a = -\frac{v_i^2}{2d} \]

  7. anonymous
    • 4 years ago
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    i see.

  8. anonymous
    • 4 years ago
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    If i have the right equation, then I must be evaluating the problem wrong. I just plugged in each variable as such: (0m/s)^2 = (20m/s)^2+2a(85m)

  9. anonymous
    • 4 years ago
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    and my final answer was 570m/s^2. Maybe I need to divide the 400 by 170....

  10. JamesJ
    • 4 years ago
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    Using the equation I wrote down, \[ a = -\frac{v^2}{2d} = -\frac{400}{170} = 2.35 \]

  11. JamesJ
    • 4 years ago
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    *dropped the negative \[ a = -2.35 \ m/s^2 \]

  12. anonymous
    • 4 years ago
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    yes I ended up with the same answer. I figured out what I did wrong thanks.

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