x = 2t^3 - 21t^2 + 60t - 30 is the motion of an ant where x is his location on the number line at time t seconds.

Question

1) When is he going to the left?
Answer: between 3 to 5 seconds

2) When is he going to the right?
Answer: between 0 to 2 seconds

3) What is his net displacement after 4 seconds?
Answer: 28 units

4) How far has he actually traveled during that time?
Answer: 50 units

Are my answers correct?

Answer 1

dx/dt = 6 t^2 - 42 t + 60

= 6 (t^2 - 7 t + 10)

= 6 (t-2)(t-5)

negative from 2 to 5 (Not 3 to 5)

Answer 2

so then would it be positive from 0 to 1?

Answer 3

I assume that we do not do negative time so

v is positive for 0<t<2 and for t >5
when t = 0, x = -30
when t = 2, x = 22
when t = 4, x = 2

so 28 from 0 to 4 (part 3)agree

part 4 +52-20 = 32

Answer 4

part 4 use absolute value of distance
is 52 + 20 = 72

Answer 5

ans part 3 is from - 30 to + 2 = +32

Answer 6

positive v from 0 to 2
negative v from 2 to 5
positive v from 5 to infinity

parabola holds water, crosses below x axis between 2 and 5
vertex at t = 3.5