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.

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.