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Debora [2.8K]
4 years ago
10

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 s

econds 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.
Mathematics
1 answer:
stepladder [879]4 years ago
6 0
When the direction is reversed  velocity = 0 

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


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