Physics question about position and time

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A man jogs at a speed of 1.2 m/s. His dog waits 1.8 s and then takes off running at a speed of 3.9 m/s to catch the man.

How far will they have each traveled when the dog catches up with the man?

Aug 24th, 2015

We can model the man's position by his speed multiplied by time. That is x = 1.2*t.

Now the dog's position has a delay, since it starts later. x_dog = 3.9*(t - 1.8) (we shift the time coordinate to account for the delay, so that at t = 1.8, the dog is at position zero).

In order to calculate how far they will have traveled when the dog catches up, we first need to find at what time they will meet. This is done by setting the two equations equal to one another:

1.2*t = 3.9*(t -1.8)

Solving for t gives:

t = 2.6 seconds.

In order to calculate the distance they've travelled, we can plug in t = 2.6 into either equation, and we'll get

x = 1.2 * 2.6 = 3.12 meters

Please let me know if you need any clarification. Always glad to help!
Aug 24th, 2015

You're a lifesaver, this was a huge help! Thank you so much for your time! This is an awesome service.

Aug 24th, 2015

Aug 24th, 2015

Would you mind helping me with this last question?

The stoplights on a street are designed to keep traffic moving at 40 mi/h. The average length of a street block between traffic lights is about 80 m.

What must be the time delay between green lights on successive blocks to keep the traffic moving continuously? There are 1.609 × 10^3 m in a mile.

Aug 24th, 2015

Unfortunately I can't answer a new question under this discussion. I'd be glad to help if you post it as a new question. Thanks!

Aug 24th, 2015

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Aug 24th, 2015
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Aug 24th, 2015
Oct 21st, 2017
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