PLEASE HELP ME UNDERSTAND 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

s = t^4 -4t^3 -8t^2 +25t

velocity = v = s' = 4t^3 -12t^2 -16t +25 = 25

4t^3 -12t^2 -16t = 0

t^3 -3t^2 -4t = 0

t(t^2-3t -4) =0

t(t-4)(t+1) = 0

set the factors = 0

t-4 = 0

t = 4 seconds for velocity to equal 25 m/s

also t=0 seconds for velocity to equal 25 m/s

also t=-1 seconds for v=25 m/s

a = v' = 12t^2 -24t -16 = 0

3t^2 -6t -4 = 0

t= 1 +/-(1/6)sqr(36+48)

= 1 +/- sqr84/6 = 1+/-2sqr21/6

= about 1+ 1.53 or -.53

= about 2.53 seconds for zero acceleration

= about 2 1/2 seconds or about -1/2 second

v(2.5) = 4(2.5)^3 -(12)2.5^2 -16)2.5 + 25

= 62.5 - 75 - 48 +25

= 87.5-123 = - 35.5 m/s

= a local and absolute minimum velocity

maximum velocity approaches infinity as time t approaches infinity

v(-1/2) = 4(-1/2)^3 -12(-1/2)^2 - 16(-1/2) +25

= -1/2 -3+8+25 = 29.5 m/s= a local minimum velocity

(ignore it, if t is restricted to non-negative numbers)

e) looks like the right answer