Here, we are required to determine how many hours your friend will drive in order to catch you.
(a)<em> Your friend will have to drive 7 and a half hours inorder to catch </em><em>you.</em>
<em>(</em><em>b)</em><em> </em><em>You </em><em>both </em><em>will </em><em>be </em><em>6</em><em>7</em><em>5</em><em> </em><em>miles</em> <em>away </em><em>from </em><em>Ellensburg</em><em> </em><em>at </em><em>that </em><em>time.</em>
If you leave at 1 pm; At 2:30pm;
- That is; 1 and a half hours after leaving; you must have covered a distance, d = 75 × 1.5
- d = 112.5miles.
Therefore, your position; S after 2:30pm is given by;
S(a) = 75t + 112.5 miles from Ellensburg.
For your friend; travelling at 90miles/hr;
- His position is given as; S(b) = 90 × t
(a) For your friend to catch you, you both must be in the same position;
75t + 112.5 = 90t
90t -75t = 112.5
t = 112.5/15
t = 7.5hours
(b) To determine how far you both are from Ellensburg; we can either evaluate:
S(b) = 90t or S(a) = 75t + 112.5
Therefore, By evaluating S(b) = 90t.
S(b) = 90 × 7.5
S(b) = 675miles from Ellensburg.
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