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The position of a particle is r(t)=4.02i^−3.0j^​+2.03k^ m.

(a) What is the velocity of the particle at t=0 s and at 1.0 s?
(b) What is the average velocity between 0 s and 1.0 s?

User Ylluminate
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Final answer:

The velocity of the particle at t = 0 s and t = 1.0 s is 0 m/s. The average velocity between 0 s and 1.0 s is also 0 m/s.

Step-by-step explanation:

The velocity of a particle can be found by taking the derivative of the position function with respect to time. In this case, the position function is r(t) = 4.02i^ - 3.0j^ + 2.03k^ m.
(a) To find the velocity at t = 0 s, we differentiate the position function by taking the derivative of each component. The derivative of 4.02i^ with respect to time is 0i^, the derivative of -3.0j^ with respect to time is 0j^, and the derivative of 2.03k^ with respect to time is 0k^. Therefore, the velocity at t = 0 s is 0 m/s.
At t = 1.0 s, we differentiate the position function again. The derivative of 4.02i^ with respect to time is 0i^, the derivative of -3.0j^ with respect to time is 0j^, and the derivative of 2.03k^ with respect to time is 0k^. Therefore, the velocity at t = 1.0 s is also 0 m/s.

(b) The average velocity between 0 s and 1.0 s can be found by subtracting the initial position from the final position and dividing by the time interval. The initial position is given as r(0) = 4.02i^ - 3.0j^ + 2.03k^ m, and the final position is given as r(1.0) = 4.02i^ - 3.0j^ + 2.03k^ m. The time interval is 1.0 s - 0 s = 1.0 s. Therefore, the average velocity between 0 s and 1.0 s is (4.02i^ - 3.0j^ + 2.03k^ m) - (4.02i^ - 3.0j^ + 2.03k^ m) / 1.0 s = 0 m/s.

User Masivuye Cokile
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