PLEASE PLEASE PLEASE HELP ME!!! PLEASE SHOW ME HOW TO DO THIS!!

Question

A particle moves with position function

s = t⁴ − 4t³ − 8t² + 25t,    t ≥ 0

A)

At what time does the particle have a velocity of 25 m/s? (Enter your answers as a comma-separated list.)

t=?

B) At what time is the acceleration 0? (Round your answer to two decimal places.)

t =?

C) What is the significance of this? (choose the correct answer)

a)At this time, the acceleration changes from negative to positive and the velocity attains its minimum value.

b)At this time, the acceleration changes from positive to negative and the velocity attains its maximum value.   

c) At this time, the acceleration changes from negative to positive and the velocity attains its maximum value.

d) At this time, the acceleration changes from positive to negative and the velocity attains a value of zero.

e) At this time, the acceleration changes from positive to negative and the velocity attains its minimum value.

Answer 1

a) Take the derivative of the position function to get the velocity function:

v(t)=4t^3-12t^2-16t+25, this function will equal 25 when the velocity is 25, so

25=4t^3-12t^2-16t+25 => 0=4t^3-12t^2-16t=4t(t^2-3t-4)=4t(t-4)(t-1)

the positive zero = 4, so at 4s the velocity is 25m/s.

b)Take the derivative of the velocity function, v(t)=4t^3-12t^2-16t+25, to get the acceleration function,

a(t)=12t^2-24t-16=4(3t^2-6t-4). When the function is equal to zero is when the acceleration will equal 0 so we need to find the positive root, which using the quadratic formula we get (3+2sqrt(21))/3.

c) a is the correct answer, we see before the positive root the function is negative, therefore the acceleration is decreasing.