The position of a certain particle depends on time according to the following equation:
\[ X(t) = t^{-5.2}t + 1.2 \]
\(X\) is in meters and \(t\) is in seconds.
A) Find the displacement and average velocity for the interval \(3.4 \, \text{s} \leq t \leq 4.5 \, \text{s}\).
Answer (3)
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 .
The question asks for the displacement and average velocity of a particle in a certain time interval in a high school Physics context. By using the position-time equation and calculating differences in position, the displacement and average velocity can be determined. ;
The displacement of the particle during the time interval from 3.4 s to 4.5 s is approximately -0.000602 m. The average velocity over this interval is approximately -0.000548 m/s, indicating it's moving in the opposite direction. Hence, the motion is in a negative directional change during this time period.
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