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gandalfwiz

okay, you have a moped constantly accelerating to get to Ke$ha

  • 2 years ago
  • 2 years ago

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  1. gandalfwiz
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    it accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. I have to find the average velocity, velocity at 5.21 seconds, and acceleration of the car.

    • 2 years ago
  2. gandalfwiz
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    I'm pretty sure that the average velocity is 21.11 m/s (110/5.210), but the other two questions are confusing me :P Any help?

    • 2 years ago
  3. Jemurray3
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    Your average velocity is correct. For the other two questions, you appear to have three unknown quantities: initial velocity, final velocity, and acceleration. This is bad, because we only have two equations: \[ v_f^2 = v_i^2 + 2a\Delta x\] \[\Delta x = v_it + \frac{1}{2} a t^2 \] But on closer inspection, since we know the time, once we find the acceleration and the initial velocity then we immediately know the final velocity, so we're okay.

    • 2 years ago
  4. Jemurray3
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    Let's do some rewriting. \[v_f = v_i + a t\] so \[v_f^2 = v_i^2 + a^2t^2 + 2v_i at\] Our first equation therefore becomes \[v_i^2 + a^2t^2 + 2v_i at = v_i^2 + 2a\Delta x\] cancelling terms, we find that \[2v_i a t = 2a\Delta x - a^2t^2\] and so \[v_i = \frac{\Delta x}{t} - \frac{1}{2}at \]

    • 2 years ago
  5. Jemurray3
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    Similarly, we find that \[v_f = \frac{\Delta x}{t} + \frac{1}{2} at\] If we plug these into the second equation, we get the following: \[\left(\frac{\Delta x}{t} + \frac{1}{2} at\right)^2 = \left(\frac{\Delta x}{t} - \frac{1}{2} at\right)^2 + 2a\Delta x \] which simplifies to \[2a^2 t^2 = 2a\Delta x\] from which we find that \[a = \frac{\Delta x}{t^2} \] We now have everything we need to determine the initial and final velocities as well as the velocity at any time t in between.

    • 2 years ago
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