Question

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

Answer 1

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.

Answer 2

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?

Answer 3

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.

Answer 4

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

Answer 5

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.