Solve the Exponential Question: Breaking Distance at 60 km/h

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SUMMARY

The discussion centers on calculating the braking distance of a car traveling at 60 km/h, given that a car traveling at 40 km/h can stop in 3 meters. The relationship between speed and braking distance is established through kinetic energy principles, where doubling the speed quadruples the braking distance. Therefore, if the speed increases by a factor of 1.5, the braking distance increases by a factor of 2.25. The expected braking distance at 60 km/h is approximately 6.75 meters.

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Exponential Question??

Homework Statement



If a car traveling at a speed of V can break at an emergency stop at a distance X. Assuming all other conditions are similar if the traveling speed of the car doubles, the stopping speed will be 4x. (I worked that out)
A driver traveling at 40 km/h in a school zone can break to emergency stop in 3 meters. What would be the breaking distance of the car were traveling at 60 km/h.

2. The attempt at a solution

I have no idea how to figure this out. From what I've done, I think the number is going to be less the 7.5 but greater than 6. I'm not exactly sure how I got that though.

Any help would be Greatly Appreciated!
 
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If you solved the first one by equating the kinetic energy to a constant braking force times distance then you understand the problem. Doubling the velocity multiplies the kinetic energy by four. I don't understand why multiplying the velocity by 1.5 is throwing you for a loop. 2^2=4, 1.5^2=?
 

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