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## gandalfwiz 3 years ago okay, you have a moped constantly accelerating to get to Ke\$ha

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1. gandalfwiz

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. gandalfwiz

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?

3. Jemurray3

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.

4. Jemurray3

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 $v_i = \frac{\Delta x}{t} - \frac{1}{2}at$

5. Jemurray3

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.

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