A particle moves along the x-axis so that its velocity at any time t > 0 is given by v(t) = 4t(t – 7)(t – 1).
A. Find the acceleration at any time t.
B. Find the minimum acceleration of the particle over the interval [0, 3].
C. Find the maximum velocity of the particle over the interval [0, 2].
+128474 A.
v(t) = 4t(t - 7) (t - 1) simplify
v(t) = 4t ( t^2 - 8t + 7)
v(t) = 4t^3 - 32t^2 + 28t
The acceleration at any time is the derivative of this = v'(t) = 12t^2 - 64t + 28
B. Set the derivative to 0 and solve for t
12t^2 - 64t + 28 = 0 divide through by 4
3t^2 - 16t + 7 = 0
The minimum acceleration on [0,3 ] occurs at t = (16)/ (2 * 3) = 16/6 = 8/3 sec
And the minimum acceleration is 3(8/3)^2 - 16(8/3) + 7 = -43/3
C. See the graph here : https://www.desmos.com/calculator/70cdiunnem
The max velocity on [0, 2 ] occurs at about .481 s and is ≈ 6.51
