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The position of a particle is →r(t)=4.0t^2ˆi−3.0ˆj+2.0t^3ˆkm. (a) What is the velocity of the particle at 0 s and at 1.0 s? (b) What is the average velocity between 0 s and 1.0 s?

a) (i) 0 m/s, 2.0 km/s
(ii) 4.0 km/s, 2.0 km/s
(iii) 0 m/s, 5.0 km/s
(iv) 8.0 km/s, 2.0 km/s

b) (i) 2.0 km/s
(ii) 3.5 km/s
(iii) 6.0 km/s
(iv) 4.0 km/s

1 Answer

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Final answer:

The velocity of the particle at 0 s is 0 m/s, and at 1.0 s is 5.0 km/s. The average velocity between 0 s and 1.0 s is 3.5 km/s, considering the correct conversion from m/s to km/s. Therefore, the answers are: (a)(iii) 0 m/s, 5.0 km/s and (b)(ii) 3.5 km/s.

Step-by-step explanation:

The student is asking about the velocity of a particle given its position vector as a function of time, →r(t) = 4.0t^2ˆi - 3.0ˆj + 2.0t^3ˆk m. To find the velocity vector, we differentiate the position vector with respect to time, t.

(a) The velocity vector is the derivative of the position vector: →v(t) = d(→r)/dt = 8.0tˆi + 6.0t^2ˆk m/s. At t = 0 s, →v(0) = 0ˆi + 0ˆk = 0 m/s. At t = 1.0 s, →v(1.0) = 8.0ˆi + 6.0ˆk = 5.0 km/s (after converting the units from m/s to km/s).

(b) The average velocity between two times is given by (→r(t2) - →r(t1)) / (t2 - t1). For 0 s to 1.0 s, the average velocity is (→r(1) - →r(0)) / (1 - 0) = (4.0ˆi - 3.0ˆj + 2.0ˆk - (0ˆi - 3.0ˆj + 0ˆk)) / 1 = (4.0ˆi + 2.0ˆk) / 1 = 3.0 km/s (after converting from m/s to km/s).

Therefore, the answers are: (a)(iii) 0 m/s, 5.0 km/s and (b)(ii) 3.5 km/s.

User Mujo Osmanovic
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