Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

The position of a certain particle depends on time according to the following equation.
X(t)=t^-5.2t+1.2
X is in meters and t is in seconds.
A) find the displacement and average velocity for the interval 3.4s less than or equal to t less than or equal to 4.5a


Sagot :

AL2006
X = t² - 5.2t + 1.2

At time 3.4s . . .
X = (3.4)² - (5.2)(3.4) + 1.2 = -4.92m

At time 4.5s . . .
X = (4.5)² - (5.2)(4.5) + 1.2 = -1.95m

The displacement over that interval is (-1.95m) - (-4.92m) = +2.97m

Average velocity = (displacement) / (time for the displacement) in the direction of the displacement.

Time interval = (4.5s - 3.4s) = 1.1 second

Average velocity = 2.97m / 1.1s = 2.7 m/s in the direction of the displacement.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.