Precalculus question

Question

This problem deals with cars traveling between Bellevue and Spokane, which are 280 miles apart. Let t be the time in hours, measured from 12:00 noon; for example, t = −1 is 11:00 am.

(a) Joan drives from Bellevue to Spokane at a constant speed, departing from Bellevue at 11:00 am and arriving in Spokane at 3:30 pm. Find a function j(t) that computes her distance from Bellevue at time t. Sketch the graph, specify the domain and determine the range.

(b) Steve drives from Spokane to Bellevue at 70 mph, departing from Spokane at 12:00 noon. Find a function s(t) for his distance from Bellevue at time t. Sketch the graph, specify the domain and determine the range.

(c) Find a function d(t) that computes the distance between Joan and Steve at time t.

Answer 1

Tricky problem.

For part a) Joan's rate is (280 miles)/(4.5 hrs) = 560/9 miles/hr and her time is t + 1 since time is measured from noon (since t = -1 at 11:00, her starting time, we want to multiply her rate times 0 and "t + 1" does that). So here distance from Bellevue would be 560/9(t + 1) or j(t) = 560/9t + 560/9

The Domain would be [-1, 3.5] and the Range would be [0, 280]

For part b), His speed is 70 so his distance traveled is 70t but since we're measuring from Bellevue, it would be 280 - 70t so s(t) = 280 - 70t

The Domain would be [0, 4] (because at 70, it takes 4 hrs to get to Bellevue) and the Range would be [0, 280]

The graphs would look like this:

The distance between them must be broken into a piecewise function:

280 - (560/9t + 560/9) For -1 < t < 0

d(t) = |(280 - 70t) - (560/9t + 560/9)| For 0 < t < 3.5 (notice the absolute value )

280 - (280 - 70t) For 3.5 < t < 4

This simplifies a bit (combining like terms for each of the 3 conditions) but you can easily do that.