Braking Distance Increase Factor w/ 50% Speed Increase

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SUMMARY

The discussion focuses on the mathematical relationship between speed and braking distance, specifically when a car's speed is increased by 50%. The conclusion is that the minimum braking distance increases by a factor of 2.25. This is derived from the kinetic energy equations, where the kinetic energy at the increased speed (KE_2) is compared to the original kinetic energy (KE_1). The key equation used is KE_1 = 1/2 m (v_1)^2 and KE_2 = 1/2 m (v_2)^2, leading to the final relationship of KE_1 = 2.25 KE_2.

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sugarntwiligh
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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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