Gerome D.

asked • 10/21/22

A point moves along a path with an equation S = 36t − 1/3 t^3 + 24 where S is in meters and t is in seconds. Find the acceleration of the point when t = 1.

S = 36t − 1/3 t^3 + 24

t = 1

1 Expert Answer

By:

Gerome D.

so if ds/dt 36t- 1/3 t^3 + 24 is 36- 1/3 (3t^2) + 0 v = 36 - 1t^2 then a =dv/dt 36-t^2 a = 0 - 2t since t = 1 -2(1) = a= -2m/s^2 is this right?
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10/21/22

Patricia D.

tutor
I said Acceleration is the derivative of S. This was written incorrectly - Acceleration is the second derivative of S. This is what you did. Your answer is correct. It would be easier to follow if you separated each step. Good job!
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10/21/22

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