Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Anybody here good at physics and curious to know how long it takes a speeding Moped to reach Ke$ha?

See more answers at
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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

The moped accelerates uniformly over a time of 5.21 seconds for a distance of 110 meters. I have to a) find the average velocity, b) find the velocity at 5.21 seconds, and c) determine acceleration of the Moped.
I'm pretty sure that the average velocity is 21.11 m/s (distance/time), but I'm unsure how to solve the others. Help?
does the moped start from rest?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I'm assuming so
My formula chart sez \[\Delta d = v_i \Delta t + \frac{1}{2}a (\Delta t)^2\] and you know d, v_initial, and t, so it's easy to solve for "a" ...
average velocity= (u+v)\2=21.21 m\s v=u+at u+v=42.22 m\s
final velocity is pretty fast, can we hope that Ke$ha won't be able to get out of the way of the speeding moped?
42.22-u=u+at 2u=42.22-5.21 a
lol @ fixer... but your formula is way too intense for me :) I think that there's a way to figure it out without one other than the basics...
ravi: what do u and v stand for?
i think ravi is saying that final velocity is twice average velocity (since initial velocity is zero), and than acceleration is (final velocity) / (time).
a=(42.22-2u)\2 u=initial velocity v=final velocity a=acceleration
hold on... let me process that information...
ravi- how'd you get 21.21? My calculations give 21.11
sir, it was written by mistake
okay, so it is 21.11? and then from there you said that velocity=initial velocity + acceleration * time?
and after that was initial velocity + final velocity= 42.22 m/s. Where did 42.22 come from?
everyone whos good at physics are at physics group
no one in the physics group :P
distance traveled=(average velocity). time=110
im glad some one could help u
20ppl there
Ravi? Where's u go :(
i am here
u+v=42.22 v-u=5.21 a so adding these equation, we get 2v=42.22+5.21a
a=(v-u)\t if moped started from rest then a=v\t=v\5.11
now we have final equation for v v=( 42.22+5.11*(v\5.11) ) \ 2 solving this equation for v, we can find the value of final velocity .
here average velocity= (u+v)\2

Not the answer you are looking for?

Search for more explanations.

Ask your own question