anonymous
  • anonymous
calculates the acceleration of a body starting from rest and has a speed of 20m / s after traveling 100m
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1342777028474:dw|
anonymous
  • anonymous
|dw:1342777061960:dw|
anonymous
  • anonymous
|dw:1342777080539:dw|

Looking for something else?

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

More answers

lgbasallote
  • lgbasallote
you are given the initial speed which is 0 (because it started from rest) you are given distance which is 100 m you are given final speed which is 20 m/s then you're looking for acceleration therefore you can use the formula \[V_2 ^2 = V_1 ^2 + 2ad\] where V2 is final speed V1 is initial speed a is acceleration d is distance so now just substitute \[20^2 = 0^2 + 2a(100)\] \[400 = 200a\] divide both sides by 200 \[\frac{400}{200} = a\] \[a = 200 m/s^2\]
lgbasallote
  • lgbasallote
uhmm oops..should be \[a = 2 m/s^2\] sorry haha
lgbasallote
  • lgbasallote
do you get it @Muskan ?
anonymous
  • anonymous
yess
lgbasallote
  • lgbasallote
glad to hear that :)
anonymous
  • anonymous
:)) thnx

Looking for something else?

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