Catch-up problem, can't seem to get it

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Homework Help Overview

The problem involves a scenario with a herd of sheep moving at a constant speed and a dog that starts at rest and accelerates. The objective is to determine the time it takes for the sheep to catch up to the dog, who has a head start of 5 meters.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to establish the positions of both the sheep and the dog as functions of time. There is an exploration of how to set these equations equal to find the meeting point. Some participants express uncertainty about the initial conditions and the correct interpretation of the problem.

Discussion Status

The discussion is ongoing, with participants providing insights into how to approach the problem by focusing on the positions of the sheep and the dog. There is a recognition of the need to clarify the roles of the different agents involved, but no consensus has been reached yet.

Contextual Notes

Participants note the lack of initial position information for the sheep and express confusion about whether to prioritize finding position or time in their calculations. There is an emphasis on the constant speed of the sheep and the accelerated motion of the dog.

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Homework Statement



I have a herd of sheep moving at 7 m/s. My dog is at rest 5m in front of the sheep and then accelerates at a rate of 3 m/s^2. how long does it take in seconds for the bulls to catch up to the cowgirl?

Homework Equations



I have a feeling that t=square root of 2s/a is relevant. But I also think that they have to be equal to each other since they meet at some point. So the V_s is 7m/s and the S_d is 5m+S_s. I know the dog's accerlation is 3m/s^2 but I don't have time here.


The Attempt at a Solution



I thought about calculating the dog's time as t=square root of 2s/a with s being 5. that gave me 1.8 seconds which i assume means 1.8 seconds ahead of the sheep but the sheep are going 7 m/s so that means that in 1.29 seconds they will catch up with the dog? Thanks for any clarification from the prospective helpers.
 
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ArmavirMig said:
I have a herd of sheep moving at 7 m/s. My dog is at rest 5m in front of the sheep and then accelerates at a rate of 3 m/s^2. how long does it take in seconds for the bulls to catch up to the cowgirl?
I assume that the problem is how long does it take for the sheep to catch up to the dog?

Measure position from the sheep's starting point. What's the sheep's position as a function of time? What's the dog's position as a function of time? Set those equal to solve for when they meet.
 
Ha, yes I wanted to replace the agents in question.

I don't have a position to start with the bulls, they are just moving, and the cowgirl is just whatever the bulls position is +5m. Should i be looking for position first and then time? I am at quite a loss.
 
ArmavirMig said:
I don't have a position to start with the bulls, they are just moving, and the cowgirl is just whatever the bulls position is +5m. Should i be looking for position first and then time? I am at quite a loss.
Assume the sheep/bulls start at x = 0. What's their position as a function of time? It's constant speed motion.

Given the above, the dog/cowgirl starts at x = 5. What's her position as a function of time? It's accelerated motion.
 

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