Tuesday

April 21, 2015

April 21, 2015

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)

physics - This is a sample question: A man jogs at a speed of 1.2 m/s. His dog ...

Math - 8*2sq+7*(4+1)= ? Can someone explain to me what to after the parenthesis...

Math - I am trying to figure out how to find the answer for r in a fraction. I ...

Math - I am trying to figure how to solve this but I am totally confused? Can ...

physics - Hi: We have a homework problem that I have no idea even how to set up...

physics - Hi: We have a homework problem that I have no idea even how to set up...

Basic Algebra - I am having a hard time trying to figure out this problem, if ...

physics - A man jogs at a speed of 1.7 m/s. His dog waits 1.9 s and then takes ...

physics - A man jogs at a speed of 1.7 m/s. His dog waits 1.4 s and then takes ...

Algebra - I am having such a hard time in this class. Hopefully someone here can...