Straight-line Kinematics problem

AI Thread Summary
The discussion revolves around a kinematics problem involving two cars traveling at 90 km/h, where the lead car brakes suddenly, and the following car delays braking. The calculations reveal that the lead car stops after 41.7 meters, while the second car, factoring in a 0.4-second delay and a slower deceleration, stops after 62.8 meters. Participants express confusion over discrepancies between their calculations and the book's answer of over 30.9 meters for the minimum separation needed to avoid a collision. One user suggests that the book's answer may be incorrect, considering its publication date and the potential for outdated methods. Ultimately, the consensus leans towards the conclusion that the book's answer does not align with their more accurate calculations.
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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