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OGrant18075
02.08.2019 •
Mathematics
The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in seconds and s(t) measured in feet. find the position and acceleration of the particle at the two instants when the particle reverses direction. include units in your answer.
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Ответ:
v = ds/dt = 6t^2 - 42t + 60 = 0
t^2 - 7t + 10 = 0
(t - 5)(t - 2) = 0
t = 2 s and 5 s which are the times when the direction is reversed
Position at t = 2 = 2(2)^3 - 21(2)^2 +60(2) + 3 = 55 feet
at t = 5 position = 28 feet
Acceleration = dv/dt = 12t - 42
At t = 2 acceleration = -18 ft s-2
At t = 5 = 18 ft s-2
Ответ: