The position of a particle moving along an x axis is given by x = 14.0t^2 - 5.00

Question

The position of a particle moving along an x axis is given by x = 14.0t^2 - 5.00t^3, where x is in meters and t is in seconds. Determine the position, the velocity, and the acceleration of the particle at t = 6.00 s. What is the maximum positive coordinate reached by the particle and at what time is it reached? What is the maximum positive velocity reached by the particle and 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 = 6.00 s

Explanation / Answer

x = 14t^2 - 5 t^3

v = 28t - 15t^2

a = 28 - 30t

part a

position at t= 6s => x =-576m

as I can see you have solved till part g

so now part h

at the instant particle is not moving ,,,,,so v = 0

we get v=  28t - 15t^2 = 0

t = 28/15 s = 1.8666 s

so at t = 1.8666s we have a = 28 - 30*1.8666 = -28 m/s2

part i

avg vel between t = 0 and t = 6s

avg vel = displacement / time

at t = 0........ x = 0 m

at t = 6s.........x = -576m

so displacement = -576m

avg vel = -576 / 6 = -96m/s

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