anonymous
  • anonymous
okay, you have a moped constantly accelerating to get to Ke$ha
Physics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.