Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
An object initially at rest experiences an acceleration of 1.7m/s^2 for 6.0s?
and then travels at that constant velocity for another 9.0s
What is the object’s average velocity over the 15s interval? in m/s
Full answer and explanation please
1 Answer
- ?Lv 71 decade agoFavorite Answer
p(t) = position function w.r.t. time
v(t) = velocity function w.r.t. time
a(t) = acceleration function w.r.t. time
a(t) = 1.7 m/s^2 t, for 0 < t < 6
For the first 6 seconds, v(t) = Integral of a(t) = 0.85 t (from 0 to 6). At t = 6, v(6) = 5.1 m/s
So, v(t) for 6 seconds = 0.85 m/s on average
v(t) for t = 6 < t < 15 = 5.1 m/s constantly (I'm assuming no friction or other forces to slow down)
average velocity = (6(0.85) + 9(5.1)) / 15
