SOLUTION: Please help me solve this problem:
An object is moving in a straight line. The distance (in meters) from a fixed point is given by s= -2t^3 - 4t^2 + 6t + 4 where t≥0. Dete
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-> SOLUTION: Please help me solve this problem:
An object is moving in a straight line. The distance (in meters) from a fixed point is given by s= -2t^3 - 4t^2 + 6t + 4 where t≥0. Dete
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Question 1006847: Please help me solve this problem:
An object is moving in a straight line. The distance (in meters) from a fixed point is given by s= -2t^3 - 4t^2 + 6t + 4 where t≥0. Determine the acceleration when the velocity is -2m/s. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! An object is moving in a straight line. The distance (in meters) from a fixed point is given by s= -2t^3 - 4t^2 + 6t + 4 where t≥0. Determine the acceleration when the velocity is -2m/s.
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s= -2t^3 - 4t^2 + 6t + 4
Velocity is the 1st derivative.
s'(t) = -6t^2 - 8t + 6
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Determine the acceleration when the velocity is -2m/s.
s'(t) = -6t^2 - 8t + 6 = -2
3t^2 + 4t - 4 = 0
(3t - 2)*(t + 2) = 0
t = 2/3 seconds (ignore t = -2)
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Accel is the 2nd derivative
s"(t) = -12t - 8
s"(2/3) = -8 -8
Accel = -16 m/sec/sec
