Final answer:
The velocity of the particle is –(t) = (6.0tî + 0.0ĵ – 6.0ĸ) m/s and the acceleration is (t) = (6.0î + 0ĵ + 0ĸ) m/s². At t = 0, the velocity is zero in all directions and the acceleration is 6.0 m/s² in the x-direction only.
Step-by-step explanation:
To find the velocity and acceleration of a particle at a generic time t, one must differentiate the position vector r(t). Given the position vector œ(t) = (3.0t²î + 5.0ĵ – 6.0tĸ) m, the velocity v(t) is the first derivative of œ(t) with respect to time and the acceleration a(t) is the second derivative.
Therefore, the calculations will be:
- Velocity: v(t) = dœ(t)/dt = (6.0tî + 0.0ĵ – 6.0ĸ) m/s
- Acceleration: a(t) = dv(t)/dt = (6.0î + 0ĵ + 0ĸ) m/s²
At time t = 0:
- Velocity: v(0) = (0 m/s, 0 m/s, 0 m/s)
- Acceleration: a(0) = (6.0 m/s², 0 m/s², 0 m/s²)