SOLUTION: A particle moves according to the parametric equations: y=2t^2 and x=t^3 where x and y are displacements(in meters) in x and y directions respectively and t is time in seconds.

Question 1145386: A particle moves according to the parametric equations: y=2t^2 and x=t^3 where x and y are displacements(in meters) in x and y directions respectively and t is time in seconds. Determine the acceleration after 3seconds.
Answer by ikleyn(52754) (Show Source): You can put this solution on YOUR website!
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Calculate y-component of the acceleration vector as the second derivative of "y" over t:    = 4 m/s^2.


Calculate x-component of the acceleration vector as the second derivative of "x" over t and substitute t = 3 seconds:    = 6t = 18 m/s^2.


Calculate the magnitude of the acceleration as  | a | =  =  =  =  m/s^2.


Use your calculator to get the numerical value.

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