Confusing Physics Question: How Far Will a Car Travel Before Overtaking a Truck?

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Homework Help Overview

The discussion revolves around a physics problem involving a car accelerating from rest and a truck moving at a constant speed. The original poster describes the scenario where the car accelerates at 6.00 m/s² while the truck travels at 21.0 m/s, seeking to determine the distance the car travels before overtaking the truck.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to convert acceleration into velocity over time and considers trial and error methods. Some participants explore algebraic approaches to find the time and distance involved in the overtaking scenario.

Discussion Status

While one participant claims to have solved the problem, providing a specific velocity and distance, there is no consensus on the correctness of these values, and the discussion remains open for further contributions or clarifications.

Contextual Notes

Participants note the potential for multiple interpretations of the problem setup and the need for clear definitions of terms such as "overtake." There is also mention of the original poster's uncertainty regarding the quickest method to approach the problem.

jen333
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Just looking over some extra physics questions and I came upon this one, it seems quite easy, but difficult when i actually attempted it:
Just as a traffic light turns green, a waiting car starts off with a constant acceleration of 6.00m/s^2. At the instant the car begins to accelerate, a truck with a constant velocity of 21.0m/s passes in the next lane. How far will the car travel before it overtakes the truck? How fast will the car be traveling when it overtakes the truck?
I attempted this question by changing the car's acceleration to a velocity by multiplying with an amount of time, etc.
I imagine that this question could possibly be done with trial and error,but I'm just wondering if there's a quicker way to attempt this question or a specific formula. Any hints or pointers, please post!
 
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problem solved!

I have solved this question, therefore i am no longer using this thread. If there is anyone who wishes to post here, by all means go ahead! :smile:
Thx,
jen333

ps) for those who would like to know, the answer is 24m/s.
 
It has been a while, so maybe I am wrong, but this is what I got:
[tex]0.5at^2=21t[/tex]
[tex]3t^2=21t[/tex]
[tex]t^2=7t[/tex]
[tex]t={0, 7}[/tex]

therefore:
[tex]v=ta[/tex]
[tex]v=42[/tex]
 
Last edited:
Yes,the velocity is 42m/s...It's correct.And the distance is 147m...

Daniel.
 

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