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Agata [3.3K]
3 years ago
7

A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

ill they have each traveled when the dog catches up with the man? Answer in units of m.
Physics
1 answer:
inessss [21]3 years ago
6 0

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is v=\frac{d}{t}, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is d=v\cdot t

Now, the distance equation for the man would be:

d_{man}=v_{man}\cdot t=1.6\cdot t

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

d_{man}=d_{dog}

1.6\cdot t=3\cdot (t-1.8)

1.6\cdot t=3\cdot t-5.4

1.4\cdot t=5.4

t=\frac{5.4}{1.4}

t=3.8571s

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

d_{man}=1.6\cdot t=1.6\cdot (3.8571)

d_{man}=6.1714m

Finally, the dog catches up with the man 6.1714m later.

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Answer:68.15m/s

Explanation:

<u><em>Given: </em></u>

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<u>Set up:</u>

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<h2><u><em>Solution:</em></u></h2><h2><u><em>68.15m/s</em></u></h2>

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