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?
Answer in units of m.
Hi there! Thank you for the opportunity to help you with your question!
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
You're a lifesaver, this was a huge help! Thank you so much for your time! This is an awesome service.
Glad to help! Please contact me if you have any other questions! Cheers
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
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.
Answer in units of s.
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!
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