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
 
 she will catch up with Tom after:
 t = 1 hour