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October 1, 2014

October 1, 2014

Posted by **John** on Saturday, August 28, 2010 at 6:03pm.

A man jogs at a speed of 1.2 m/s. His dog waits 2 s and then takes off running at a speed of 4.1 m/s to catch the man. How far will they each have traveled when the dog catches up with the man?

The answer is 3.3931 m.

I just need to know the process to solve this practice problem so I can use that to solve my actual problem. I don't want to post my actual problem because I feel that could be akin to cheating if someone answered it.

Thank you in advance!

I felt like I should set the two equations equal to one another (since at a certain time their distances will be equal) but I'm failing to come up with a workable equation or process.

- physics -
**MathMate**, Saturday, August 28, 2010 at 6:31pmIf you let t be the time in seconds the man started jogging.

then (t-2) is the duration the dog ran.

The distance travelled by both are the same, so calculate the distance travelled by the man and the dog. Equate them and solve for t.

Substitute t back into*each*of the distance expressions to make sure they are equal.

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