Asked by fareha
                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?
            
            
        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?
Answers
                    Answered by
            Reiny
            
    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.
    
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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