Question

Two cars travel in the same direction along a straight highway, on at a constant speed of 38.0 mi/h and the other at 89.0 mi/h. (a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 44.0 mi away? (b) How far must the faster car travel before it has a 14.0-min lead on the slower car? (The faster car is far enough ahead, that it will take the slower car 14.0-min to cover this distance.)
a. 38 * t1 = 44 mi.
Solve for t1.

89*t2 = 44. Solve for t2.

T = t1-t2

b. d = 38(14/60) = 8.9 mi.

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