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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