x = 16 * t^2 – 6 * t^3

x = 16 * 3^2 – 6 * 3^3 = -18 meters

To determine the equation of velocity versus time, take the first derivative of the equation above.

vf = vi + 32 * t – 18 * t^2, vi = 0

vf = 32 * t – 18 * t^2

vf = 32 * 3 – 18 * 3^3 = -66 m/s

The equation of acceleration versus time is the first derivative of this equation.

a = 32 – 36 * t

a = 32 – 36 * 3 = -7.6 m/s

The maximum positive velocity is when the acceleration is 0.

a = 32 – 36 * t = 0

t = 8/9 second

v = 32 * 8/9 – 18 * (8/9^2

This is 14 2/9th m/s.

(h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)?

Let’s use the following equation to determine the when the velocity is 0 m/s

vf = 32 * t – 18 * t^2

0 = 32 * t – 18 * t^2

18 * t^2 = 32 * t

t = 32/18

This is 1 7/9th second.

a = 32 – 36 * 32/18 = -32 m/s

i) Determine the average velocity of the particle between t = 0 and t = 3.00 s.

At 3seconds, the velocity is -66 m/s.

v ave = ½ * (vi + vf), vi = 0

v ave = -33 m/s

# The position of a particle moving along an x axis is given by x = 16.0t2 - 6.00t3, where x is in meters and t is in seconds.?

The position of a particle moving along an x axis is given by x = 16.0t2 - 6.00t3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 3.00 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (i) Determine the average velocity of the particle between t = 0 and t = 3.00 s.

## 1 Answers

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