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.

At a certain time a particle had a speed of 80 m/s in the positive x direction, and 7.0 s later its speed was 90 m/s in the opposite direction. What was the average acceleration of the particle during this 7.0 s interval? I can think of two answers that I could get: 24.29m/s^2 1.43 m/s^2 Any help would be appreciated.

Physics
See more answers at brainly.com
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

SIGN UP FOR FREE
I got 24.29 too but I haven't done these kind of problems in a long time. I just looked at the change of speed ( so from 80 to 0 and 0 to 90 considering it is in the opposite direction). So I put the acceleration as 24.28. ^^ ( well a = speed / delta t
or if you want to spend the time on it, draw the graph, find the rule of the velocity graph ( should be a linear one ax+b) and integrate it ^^
Ok, thanks! I just wanted another set of eyes to look at it.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question