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, we need to find the second derivative of the position vector (x, y) with respect to time.
Given that x = 2t^3 and y = 3t^2, we can differentiate these equations with respect to time to find the velocity components:
dx/dt = 6t^2
dy/dt = 6t
Then, differentiating these again, we can find the acceleration components:
d^2x/dt^2 = 12t
d^2y/dt^2 = 6
Now we can plug in t = 2 to find the acceleration of the car at that time:
d^2x/dt^2 = 12(2) = 24 m/s^2
d^2y/dt^2 = 6 m/s^2
So, at t = 2 seconds, the acceleration of the car is 24 m/s^2 in the x-direction and 6 m/s^2 in the y-direction.
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