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# Can you answer this calculus question?

A particle starts at time t=0 and moves along the x-axis so that its position at any time t≥0 is given by x(t)=(t-1)^3*(2t-3).

a) For what values of t is the velocity of the particle less than zero?
b) Find the value of t when the particle is moving and the acceleration is zero.

The velocity is (t-1)^2*(8t-11).

A particle starts at time t=0 and moves along the x-axis so that its position at any time t≥0 is given by
x(t)=(t-1)^3*(2t-3).

a) For what values of t is the velocity of the particle less than zero?

v(t) = x’(t) = 3(t-1)^2*(2t-3)+2(t-1)^3
= (t-1)^2*(3*(2t-3)+2(t-1))
= (t-1)^2*(6t-9+2t-2)
= (t-1)^2*(8t-11)

v(t) has odd degree (3), so end behaviors are opposite.

t → ∞ ==> v(t) → ∞

end behavior is up to the right and down to the left

zeros are 1, with multiplicity 2, and 11/8=1+3/8

going from right to left, v in Quadrant I comes down and crosses x-axis at 11/8, goes down to a turning point and back up to touch and bounce off x-axis at 1, then goes in negative direction forever.

–––––0––––––0++++> v(t)
|––––|––––––|––––> t
0      1        11/8

So v(t) < 0 for x < 1 and 1 < x < 11/8.
In interval notation: (–∞,1)U(1,11/8)

b) Find the value of t when the particle is moving and the acceleration is zero.

v(t) = (t-1)^2*(8t-11).

a(t) = v’(t) = 2(t-1)(8t-11)+8(t-1)^2

= 2(t-1)((8t-11)+4(t-1))

= 2(t-1)(12t-15)

= 6(t-1)(4t-5)

a(t) is a parabola opening up

with zeros at t=1 and t=5/4=1+2/8.

+++++0––––––0++++++++++++> a(t)
–––––0––––––––––––––0++++> v(t)
|––––|––––––|–––––––|––––> t
0      1      (1+2/8)   (1+3/8)

At t = 5/4 = 1+2/8 the particle is moving, v(1+2/8) ≠ 0, and the acceleration is zero, a(1+2/8) = 0.
V(t) = X' ( t) = 3( t - 1) ^2 ( 2t -3 ) +2t ( t- 1) ^3

= ( t - 1) ^2 ( 6t - 9 + 2t -2) =

= ( t -1 ) ^2 ( 8t -11)

8t < 11 ,  t < 11/8

a( t) = v'(t) = X" ( t) = 2( t-1) ( 8t - 11 ) + 8 ( t - 1) ^2

= 2 ( t -1) ( 8t -11 +4t -4 ) =

= 2( t -1) ( 12t - 16) =
= 8t ( t-1) ( 3t -4 )

a( t) < 0       1 < t < 4/3