Calculating Distance Between Stop Signs Using Accelerated Motion Formulas

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Homework Help Overview

The discussion revolves around calculating the distance between two stop signs using principles of accelerated motion. The scenario involves a car that accelerates, coasts, and then decelerates, prompting the need for kinematic equations to determine the total distance traveled.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations and the need to break the problem into three distinct intervals. There are attempts to calculate distances for each interval based on acceleration and deceleration, while some participants question the correctness of the initial calculations and the application of formulas for accelerated motion.

Discussion Status

The discussion is ongoing, with some participants providing calculations for the first two intervals while expressing uncertainty about the third interval. There is a suggestion to reconsider the formulas used for accelerated motion, indicating a productive direction for further exploration.

Contextual Notes

Participants are navigating the complexities of applying kinematic equations correctly, with an emphasis on understanding average velocity in the context of accelerated motion. There is a recognition that the initial approach may not fully align with the principles of motion being discussed.

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A car starts from rest at a stop sign. It accelerates at 4.20m/s^2 for 7.10s, coasts for 2.10s, and then slows down at a rate of 3.10m/s^2 for the next stop sign. How far apart are the stop signs?

Please help me start this problem. Not sure where to begin. Thanks
 
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1st Interval

Vf= 0 + 4.20m/s^2(7.10s)
= 29.82m/s

29.82m/s x 7.10s
= 211.72m

2nd Interval

29.82m/s x 2.10s
= 62.622m

3rd Interval

Vf = 29.82m/s - 3.10m/s^2(t)
-29.82m/s = -3.10m/s^2(t)

9.62s = t

29.82m/s - 3.10m/s^2(9.62s)
??

Well, do I have the first two intervals correct?

And I am stuck on the third interval... Does it just equal 29.82m, which means the distance between the two stop signs is 211.722+62.622+29.82= 304.164m

Any help would be greatly appreciated.
 
interval 1 is not right...

Remember this is accelerated motion... d = vt doesn't work here unless by v you mean "average velocity"...

what displacement formulas do you have for accelerated motion...

try to apply one here...
 

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