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Find the speed of a particle at t = 3 seconds if its acceleration is given by dv/dt = 9t^2 - 3t^3.

1. Speed v = dx/dt = 18t - 3t^2
2. Maximum speed occurs when dv/dt = 0
3. At t = 3, speed is maximum
4. x_max = 81 - 27 = 54 m

Correct answer: At t = 3, speed is maximum

Solution

The acceleration dv/dt = 9t^2 - 3t^3 becomes zero when t = 3 seconds, indicating that the speed is maximum at this time.

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