The position of a particle moving along an x axis is given by x = 17t^2 - 6.0t^3

Question

The position of a particle moving along an x axis is given by x = 17t^2 - 6.0t^3, where x is in meters and t is in seconds. Determine the position, velocity, and acceleration of the particle at t = 3.0 s. x = v = a = What is the maximum positive coordinate reached by the particle? At what time is it reached? What is the maximum positive velocity reached by the particle? At what time is it reached? What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? Determine the average velocity of the particle between t = 0 and t = 3 s.

Explanation / Answer

x= 17t^2-6t^3

= 17 ( 3)^2 - 6 ( 3)^3

=-9 m

x = 17t^2- 6t^3

0 = 17 t^2 - 6t^3

t = 2.83

x = 17( 2.83)^2 - 6( 2.83)^3 = 0.16 m

-----------------------------------------------------------------------------------------------------------------

v = dx/dt = d/dt ( 17 t^2- 6t^3) = 34t - 18 t^2

dv/dt = 34t- 18t^2

= 34-36t

0 = 34- 36t

t = 10.94

v(34/36)=289/18 m/s 16.06 m/s

------------------------------------------------------------------------------

average velocity, we want: v avg = x(3) -x(0) / 3

                                                       = 17( 3)^2 - 6(3)^3 - 0/3

                                                       = -3 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.