SOLUTION: 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 veloci

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Question 1006618: 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 ikleyn(52754)   (Show Source): You can put this solution on YOUR website!
.
The procedure is as follows:

1. Take the derivative of s by t.
   You will get a quadratic polynomial of t for the velocity, v(t) = .

2. Solve the quadratic equation v(t) = -2 to find the time moment when the velocity is - 2 .

3. Take the second derivative of s by t to get the expression for the acceleration a(t).

4. Substitute the found value of t into the expression for acceleration to determine the unknown acceleration.


Good luck!


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