Question

The position of a particle for t > 0 is given by →r(t)=(3.0t^2ˆi−7.0t^3ˆj−5.0t^−2ˆk)m. (a) What is the velocity as a function of time? (b) What is the acceleration as a function of time? (c) What is the particle’s velocity at t = 2.0 s? (d) What is its speed at t = 1.0 s and t = 3.0 s? (e) What is the average velocity between t = 1.0 s and t = 2.0 s?

a) (i) v(t)=(6.0tˆi-21.0tˆ2ˆj+10.0t^−3ˆk)m/s
(ii) v(t)=(6.0tˆi-21.0t^−2ˆj+10.0tˆ−3ˆk)m/s
(iii) v(t)=(6.0tˆi-21.0tˆ3ˆj+10.0tˆ−2ˆk)m/s
(iv) v(t)=(6.0tˆi-21.0tˆ−3ˆj+10.0tˆ−2ˆk)m/s

b) (i) a(t)=(6.0i-42.0tˆj-20.0t^−3ˆk)m/s²
(ii) a(t)=(6.0i-42.0t^2ˆj-20.0t^−2ˆk)m/s²
(iii) a(t)=(6.0i-42.0tˆ3ˆj-20.0t^−2ˆk)m/s²
(iv) a(t)=(6.0i-42.0tˆ−2ˆj-20.0tˆ−3ˆk)m/s²

c) (i) 12.0 m/s
(ii) 18.0 m/s
(iii) 6.0 m/s
(iv) 24.0 m/s

d) (i) 13.0 m/s, 27.0 m/s
(ii) 11.0 m/s, 25.0 m/s
(iii) 12.0 m/s, 26.0 m/s
(iv) 10.0 m/s, 24.0 m/s

e) (i) 18.5 m/s
(ii) 17.0 m/s
(iii) 19.5 m/s
(iv) 16.0 m/s

Answer 1

The velocity of a particle is the first derivative of its position function, and the acceleration is the second derivative of the velocity function. The correct velocity as a function of time for the given position function is v(t)=(6.0tˆi-21.0t²ˆj+10.0t⁻³ˆk)m/s, and the correct acceleration function is a(t)=(6.0i-42.0tˆj-20.0t⁻´ˆk)m/s².

Step-by-step explanation:

The position of a particle is given by the equation →r(t)=(3.0t²ˆi−7.0t³ˆj−5.0t⁻²ˆk)m. To find the velocity and acceleration as functions of time, derivatives of the position function with respect to time are taken. The velocity function, v(t), is the first derivative of the position function, and the acceleration function, a(t), is the second derivative of the velocity function.

(a) The correct velocity function is v(t)=(6.0tˆi-21.0t²ˆj+10.0t⁻³ˆk)m/s (option iii), which is obtained by differentiating the position function component-wise.

(b) The correct acceleration function is a(t)=(6.0i-42.0tˆj-20.0t⁻´ˆk)m/s² (option ii), which is obtained by differentiating the velocity function component-wise.