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
Finding the maximum distance?
I had a physics question that asked me to find the maxium X value, where X is in meters.
I was first asked to find meters traveled, then velocity, then acceleration for the function X= 16t^2 - 6.0t^3 after 3 seconds. I found those, by plugging in 3 for t and then taking the first and second derivatives. Which wasn't a problem, but then using the same function I needed to find Xmax and Vmax.
I found that Xmax was something after 1.75 seconds, which my online homework said I got right, but in order to find it, I was just plugging in numbers, but I don't know how to actually find it, and my textbook is silent on the matter.
Any help would be greatly appreciated.
1 Answer
- Anonymous1 decade agoFavorite Answer
Xmax is determined by taking the derivative of X with respect to T and setting it equal to 0.
(dX/dt) = 32t - 18t^2 = 0
Hence, t = 32/18 = 1.78 sec.
This means that at t = 1.78, the body has attained Xmax, i.e.,
Xmax = 16(1.78^2) - 6(1.78^3)
Xmax = 16.86 meters
To get the Vmax, get the second derivative of X,
(d^2X/dt^2) = 32 - 36t
and setting this equal to zero, you get t = 32/36 = 0.888 sec.
Therefore,
Vmax = 32(0.888) - 18(0.888^2)
Vmax = 15.80 m/sec.