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September 16, 2014

September 16, 2014

Posted by **j** on Saturday, January 15, 2011 at 12:55pm.

A man jogs at a speed of 1.8 m/s. His dog

waits 2 s and then takes off running at a speed

of 4 m/s to catch the man.

How far will they have each traveled when

the dog catches up with the man?

Correct answer: 6.54545 m.

I think that I'll need to set the two equations equal and that I see that I'm given time and two sets of constant velocity but I looked at all my equations that I have and I can't figure out what equation I need.

Any help would be welcome.

- physics -
**Marth**, Saturday, January 15, 2011 at 2:13pmV=d/t

d=V*t

Let t be the time the man is jogging. t-2 is the time the dog is running (since the dog waits 2s). When they catch up with each other displacement will be equal.

(4m/s)(t-2)=(1.8m/s)(t)

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