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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



Answer :

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.

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