Where Does Aaron Catch Up to Alyssa?

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Homework Help Overview

The problem involves two individuals, Alyssa and Aaron, where Alyssa travels at a constant velocity while Aaron accelerates after a brief hesitation. The objective is to determine the point at which Aaron catches up to Alyssa.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster considers the time of meeting to be the same for both individuals and questions whether to include the initial hesitation in their calculations. Some participants clarify that while the distances are equal, the times taken differ due to Aaron's delay.

Discussion Status

Participants are actively discussing the relationship between distance and time for both individuals. There is a recognition of the need to equate the distances traveled, and some guidance has been offered regarding the setup of the equations. However, there is no explicit consensus on the correct approach yet.

Contextual Notes

There is a mention of the importance of unit consistency, and participants are navigating the implications of Aaron's initial delay on the overall timing of the problem.

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hey,

I was given a problem that states:

Alyssa passes by Aaron at a constant velocity of 80km/h. Aaron wanting to catch up to her hesitates for 1 second and then starts accelerating at 10km/h/s. Where does he catch up with alyssa?

OK what i have been thinking and trying out is that the time they meet will be the same right? like when you sub the numbers into an equation the time you get is the same for both? so then you can use that time and figure out the distance, since that's the same too. ALso, so when i get the time that's it right? i don't add the 1 second to it?? the time i got was 14.8 seconds and a distance of 328.9 meters. Did i do this right or am i way off??

thanx for your time.
 
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The distance is the same but the time is not same.
Time taken by Aaron in moving the same distance is one second less then taken by Alyssa.
Pay attantion to the units!
 
D of the constant velocity person is = 80T
the accelerating person's D = .5A(T-1)^2

you can make those equal to each other and solve for T
 
Last edited:
Uhh, Moose, Distance= 80T
 
HallsofIvy said:
Uhh, Moose, Distance= 80T
:blushing: :redface:
oops :bugeye:
Changed it, thanks!
 

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