Straight-line Kinematics problem

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Homework Help Overview

The problem involves two automobiles moving at the same speed, where the lead car applies brakes and the following car has a delayed reaction. The goal is to determine the minimum separation required to avoid a collision based on their deceleration rates.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different methods for calculating stopping distances and question the validity of their results compared to the textbook answer. Some express confusion about discrepancies in their calculations.

Discussion Status

Multiple interpretations of the problem are being explored, with participants sharing their calculations and questioning the accuracy of the textbook answer. Some guidance has been offered, but no consensus has been reached regarding the correct minimum separation distance.

Contextual Notes

There is mention of the textbook being published in the pre-calculator era, which raises questions about the reliability of its calculations compared to modern methods.

LD_90
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I found this problem in the text "Elementary Physics Classical and Modern" by Weidner and Sells.

Two automobiles are both moving at 90 km/h in the same direction, one directly behind the other. The driver of the lead car suddenly applies his brakes, decelerating at 7.5 m/s^2. The other driver applies his own brakes after a delay of 0.40 s, and slows down at a rate of 6.0 m/s^2. (His tires are worn.) If there is to be no collision, what is the minimum separation between the cars at the instant the lead car's brakes are applied?

The answer given in the back of the book is "greater than 30 5/6 m or about 30.9 m." I'm wondering if this anwer is correct.

Let the lead car be car A: vo=25m/s , v=0 , a=-7.5m/s^2
using the formula v^2=vo^2+2a(x-xo) , car A stops after 41.7m, car B after 52.8m+10m(for the delay). I can get a close anwer by using a different approach. Why doen't the first method give the correct answer? :confused:
 
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LD_90 said:
Let the lead car be car A: vo=25m/s , v=0 , a=-7.5m/s^2
using the formula v^2=vo^2+2a(x-xo) , car A stops after 41.7m, car B after 52.8m+10m(for the delay). I can get a close anwer by using a different approach. Why doen't the first method give the correct answer? :confused:
This method seems correct to me. I get a minimum lead of 20.4m to avoid a collision. What other method did you use?
 
Thanks Doc Al. That's the answer that I got too. Since it did not match the back of the book answer, I took the difference in stopping times, (plus the delay) and multiplied it by the initial velocity. This gives about 30.8m. This gets a result closer to the book answer, but it doesn't seem right to me. Oh by the way, the book was published pre-calculator in 1973 with the original in the mid sixties, so I'm not sure how far slide-rule results tend to be from calculator results.
 
LD_90 said:
Since it did not match the back of the book answer, I took the difference in stopping times, (plus the delay) and multiplied it by the initial velocity.
That makes no sense to me.
Oh by the way, the book was published pre-calculator in 1973 with the original in the mid sixties, so I'm not sure how far slide-rule results tend to be from calculator results.
Come on, things weren't that bad back then! :rolleyes:
 
Well I guess the answer in the book is wrong. Thanks for the help.
 

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