SOLUTION: At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads a

Question 95345This question is from textbook prentice hall mathematics algebra 1
: At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads after Tom. If she travels at a rate of 30 miles per hour, when will she catch up?
I don't understand how to solve the problem. I dont even know where to begin. Could you please help with steps and tips of how to solve this problem? This question is from textbook prentice hall mathematics algebra 1

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads after Tom. If she travels at a rate of 30 miles per hour, when will she catch up?
:
Let t = T's travel time to when M catches up
:
Since M left 10 min later we can say: (change 10 min to hrs 10/60 = 1/6)
(t-(1/6) = M's travel time
:
One thing we do know is that when M catches up with T they will have traveled
the same distance. (This applies to all these "catch-up" problems)
:
That means we can make a distance equation:
Dist = speed * time
:
M's dist = T's dist
30(t-(1/6)) = 25t
30t - (30/6) = 25t
30t - 5 = 25t
30t - 25t = +5
5t = 5
t = 1 hr + 1:30 = 2:30 Pm when M catches up with T
:
we can check this to see if they do, indeed, travel the same distance
in the time we found. M's time = (5/6) hr and T's time = 1 hr
30*(5/6) = 25 mi
25 * 1 = 25 mi
:
Did this make sense to you? Any questions?

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