Question

The position of a particle is given by the following expression, where t is time measured in seconds: r(t) = [(3.33 m/s²)t²]i + (-4.11 m)j + [(2.18 m/s³)t³]k. What is the magnitude of the velocity of the particle, in m/s, at t = 0.00 s?

Answer 1

The magnitude of the velocity of the particle at t = 0.00 s is 0 m/s.

Step-by-step explanation:

The position of a particle is given by the equation r(t) = [(3.33 m/s²)t²]i + (-4.11 m)j + [(2.18 m/s³)t³]k.

To find the magnitude of velocity at t = 0.00 s, we need to differentiate the position function with respect to time to find the velocity function.

The velocity function, v(t), is obtained by taking the derivative of r(t). After differentiating, we get v(t) = (6.66 m/s²t)i + 0j + (6.54 m/s³t²)k.

Now, to find the magnitude of velocity at t = 0.00 s, we substitute t = 0.00 s into the velocity function. This gives us v(0.00 s) = (6.66 m/s² * 0)i + 0j + (6.54 m/s³ * 0²)k = 0i + 0j + 0k = 0 m/s.

Therefore, the magnitude of the velocity of the particle at t = 0.00 s is 0 m/s.