A particle moves according to the law of motion......

Question

A particle moves according to the law of motion s= t^3 - 8t^2 + 3t, t ≥ 0,

where t is measured in seconds and s in feet.

a) find the velocity at time t.

b) what is the velocity after 3 seconds?

These two I've got the answers, however, I don't know how to calculate the following questions.

c) When is the particle at rest? Enter your answer as a comma separated list. Enter None if the

particle is never at rest.

d) When is the particle moving in the positive direction?

    When  0 ≤t< [answer] and t> [answer]

I really need help on these questions, please help

Answer 1

a) if s is distance traveled, then s'=v=velocity=3t²-16t+3. just take the first derivative

b) At t=3, v=3(3)²-16(3)+3=27-48+3=-18 feet per second That's negative velocity, going backwards

c) At rest would mean velocity is zero, set 3t²-16t+3=0 and solve for t. It's a quadratic equation, with no obvious factors. Use either the quadratic formula or complete the square. Also maybe plug in a few easy integers to see the general range for the answer. At t=0 v=3. at t=1 v=-10 Somewhere between t=0 and t=1 the velocity will be zero. It looks like a little more than t=0, maybe t=1/4 or so. Check that by plugging t=1/4 into the quadratic equation to get: 3(1/16)-16(1/4)+3=3/16-64/16+48/16=-13/16. It's still negative velocity. For v=0, it must be even closer to t=0, a smaller fraction, maybe t=1/6. 3(1/36)-16(1/6)+3= 3/36-96/36+108/36=+15/36. The correct answer is somewhere between t=1/4 and t=1/6

Using the quadratic formula, t=(-b + or - sqr(b²-4ac)/2a t=(16 + or - sqr(256-36))/6 =8/3 - (220)^1/2/6 = approximately 0.195 or about 1/5. there is also a 2nd solution by adding 16 + sqr(220) all over 6 which equals 30.83/6 = 5.138 seconds. the particle has zero velocity at t=0.195 and at t=5.138

t=0.195, 5.138. Quadratic equations have 2 solutions, as in this problem, although sometimes the solutions repeat or or imaginary.

d) It has positive velocity when t is less than 0.195 seconds. v is positive between t=0 and about t=0.195 seconds and positive velocity between t=5.138 seconds and infinity. Or the two intervals [0,0.195] U [5.138, infinity) There's only one interval where the velocity is negative, [0.195, 5.138]. Elsewhere, it's positive.