A car is traveling along a curved road described by the parametric equations x = 2t^3 and y = 3t^2 where x and y are in meters and t is in seconds . Find the acceleration of the car when t = 2
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To find the acceleration of the car when t = 2, we need to find the second derivative of both x and y with respect to t.
First, we find the first derivative of x and y:
dx/dt = d(2t^3)/dt = 6t^2
dy/dt = d(3t^2)/dt = 6t
Next, we find the second derivative of x and y:
d^2x/dt^2 = d(6t^2)/dt = 12t
d^2y/dt^2 = d(6t)/dt = 6
Finally, we can substitute t = 2 into the second derivatives to find the acceleration:
a_x = d^2x/dt^2 = 12(2) = 24 m/s^2
a_y = d^2y/dt^2 = 6
Therefore, the acceleration of the car when t = 2 is 24 m/s^2 in the x-direction and 6 m/s^2 in the y-direction.
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