Posted by **fareha** on Monday, December 5, 2011 at 6:46am.

5. A particle moves along the x-axis in such a way that its position at time t is given by x=3t^4-16t^3+24t^2 for -5 ≤ t ≤ 5.

a. Determine the velocity and acceleration of the particle at time t.

b. At what values of t is the particle at rest?

c. At what values of t does the particle change direction?

d. What is the velocity when the acceleration is first zero?

- calculus -
**Reiny**, Monday, December 5, 2011 at 9:10am
a) velocity = x' = 12t^3 - 48t^2 + 48t

acc. = x'' = 36t^2 - 96t + 48

b) to be at rest, x' = 0

12t^3 - 48t^2 + 48t = 0

12t(t^2 - 4t + 4) = 0

12t(t-2)^2 = 0

t = 0 or t = 2

c) mmmh, what must have happened to the velocity when direction is changed?

d) for acc. = 0, 36t^2 - 96t + 48 = 0

3t^2 - 8t + 4 = 0

(t-2)(3t - 2) = 0

t = 2 or t = 2/3

so when t = 2/3, sub that into x'

I will let you do the arithmetic.

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