Braking Distance Increase Factor w/ 50% Speed Increase

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Increasing a car's speed by 50% results in a minimum braking distance increase factor of 2.25, assuming all other conditions remain constant. The discussion revolves around deriving this factor using kinetic energy equations. By expressing the relationship between initial and final speeds mathematically, participants clarify how kinetic energy changes with speed. The key takeaway is that the braking distance is proportional to the square of the speed increase. Understanding this relationship is crucial for grasping the physics of braking distances.
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Homework Statement



If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time.

Homework Equations




.5(2m)(v1+5.0)^2 = .5m(v2+5.0)^2

I know the answer is 2.25 and is deduced from the above equation. I tried to simplify:

v1^2+10v1+25 = .5v2^2+5v2+12.5

but I am still confused. I know I should eliminate time as a variable, but I am unsure of how to do this.
 
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sugarntwiligh said:
.5(2m)(v1+5.0)^2 = .5m(v2+5.0)^2
I don't understand what this equation represents.

Try this:

KE_1 = 1/2 m (v_1)^2

KE_2 = 1/2 m (v_2)^2

If v_2 is 50% greater than v_1 (express that mathematically), how does KE_2 compare with KE_1?
 
So,
(v_1) = (v_2) + (v_2 * 0.5) = 1.5 (v_2)
and, if I substitute KE_1 and KE_2 into the equation,
I get KE_1=2.25KE_2.
OMG THANK YOU!
 
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