At 1:30 PM Tom leaves his boat from a dock and head south. He travels at a rate of 25 mph. 10 minuets later, Mary leaves the same dock in her speedboat and heads after Tom. If she travels at a rate of 30 mph, when will she catch up with Tom?
We write the cinematic equation for each case. We have then: For tom: r = 25t For Mary r = 30 (t- (10/60)) We observe that we have a system of two equations with two unknowns. Solving for t we have: t = 1 hour Answer: she will catch up with Tom after: t = 1 hour
