Question

Erin travels north and south from Main Station. The distance, in km, of the train from Main Station is

modeled by the function d(t) = t3 - 9t2 + 6t, where North is positive and South is negative. Time
elapsed after the start of a shift, in hours, is represented by t, where t € (0,12]. 'If the shift starts at
noon, determine at which time(s) the train is more than 16 km south of Main Station.

Answer 1

The times are t = 9, t = 10, t = 11 and t = 12

Explanation:

For the train to be more than 16 km South and since south is taken as negative,

d(t) > -16

t³ - 9t² + 6t > -16

t³ - 9t² + 6t + 16 > 0

Since -1 is a factor of 16, inserting t = -1 into the d(t), we have

d(-1) = (-1)³ - 9(-1)² + 6(-1)+ 16 = -1 - 9 - 6 + 16 = -16 + 16 = 0. By the factor theorem, t + 1 is a factor of d(t)

So, d(t)/(t + 1) = (t³ - 9t² + 6t + 16)/(t +1) = t² - 10t + 16

Factorizing t² - 10t + 16, we have

t² - 2t - 8t + 16

= t(t - 2) - 8(t - 2)

= (t -2)(t - 8)

So t - 2 and t - 8 are factors of d(t)

So (t + 1)(t -2)(t - 8) > 0

when t < -1, example t = -2 ,(t + 1)(t -2)(t - 8) = (-2 + 1)(-2 -2)(-2 - 8) = (-1)(-4)(-10) = -40 < 0

when -1 < t < 2, example t = 0 ,(t + 1)(t -2)(t - 8) = (0 + 1)(0 -2)(0 - 8) = (1)(-2)(-8) = 16 > 0

when 2 < t < 8, example t = 3 ,(t + 1)(t -2)(t - 8) = (3 + 1)(3 -2)(3 - 8) = (4)(1)(-5) = -20 < 0

when t > 8, example t = 9,(t + 1)(t -2)(t - 8) = (9 + 1)(9 -2)(9 - 8) = (10)(7)(1) = 70 > 0

Since t cannot be negative, d(t) is positive in the interval 0 < t < 2 and t > 8

Since t ∈ (0, 12]

In the interval 0 < t < 2 the only value possible for t is t = 1

d(1) = t³ - 9t² + 6t = (1)³ - 9(1)² + 6(1) = 1 - 9 + 6 = -2

Since d(1) < -16 this is invalid

In the interval t > 8 , the only possible values of t are t = 9, t = 10.t = 11 and t = 12.

So,

d(9) = 9³ - 9(9)² + 6(9) = 0 + 54 = 54 km

d(10) = 10³ - 9(10)² + 6(10) = 1000 - 900 + 60 = 100 +60 = 160 km

d(11) = 11³ - 9(11)² + 6(11) = 1331 - 1089 + 66 = 242 + 66 = 308 km

d(12) = 12³ - 9(12)² + 6(12) = 1728 - 1296 + 72 = 432 + 72 = 504 km