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