Answer :
Final answer:
The position of the particle at t=6s is -216 meters, the velocity is -324 m/s, and the acceleration is -152 m/s^2.
Explanation:
The equation given, x = 14.0t^2 − 5.00t^3, defines the position of a particle moving along a one-dimensional path or the x-axis.
- Position: Substituting the provided time into the position equation gives x = 14(6^2) - 5(6^3) = -216 meters.
- Velocity: The velocity of the particle at any time 't' can be found by taking the first derivative of the position equation with respect to time. The first derivative of x = 14.0t^2 − 5.00t^3 is v = 28.0t - 15.00t^2. Substituting t = 6 gives v = 28*6 - 15*(6^2) = -324 m/s.
- Acceleration: The acceleration at any time 't' could be found by taking the first derivative of the velocity equation or the second derivative of the position equation with respect to time. The first derivative of v = 28t - 15t^2 is 28 - 30t. Therefore, substituting t = 6 gives acceleration = 28 - 30*6 = -152 m/s^2.
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