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The task goes as follows.

When we get our drivers-permit we learn that the braking length grows proportional to the speed squared. (If we double the speed, the braking distance is quadrupled. Assume that the road is leveled and that the braking force is constant.)

When two quantities are proportional I thought that implies that if you divide the first quantity with the other you should get roughly the same number for different numbers. The expression under should be a constant.

$$\dfrac{\text{Braking Length}}{\text{Speed}^2}$$

Is this interpretation correct?

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My heart is broken. – user21436 Mar 11 '12 at 13:39
@KannappanSampath, Why? – Inquest Mar 11 '12 at 13:50
When your brakes break, then,... – user21436 Mar 11 '12 at 13:53
My hart just slowed down. That was good because my car's brakes meant it didn't break, and so neither did the hart! (Although I must admit that after the encounter, I thought my heart was nearly broken, it was pounding so hard.) – Neal Mar 11 '12 at 14:02
up vote 1 down vote accepted

Yes. "Proportional" means exactly that the ratio of the two is constant. Alternatively, there's some number $C$ such that $brakingLength = C\times (speed)^2$.

As you are no doubt studying in your physics text, if you are applying a constant braking force $F$ to an object (car), then over the distance that force is applied the work done by the braking force is equal to the change in kinetic energy of the object. If $L$ is the braking length, the distance you travel until you stop, and $M$ is the mass of your object $FL = \frac{1}{2}Mv^2$, so $$L = \bigg(\frac{M}{2F}\bigg)v^2.$$ The factor in parentheses is the constant $C$ since $M$ and $F$ are constants. Alternatively, $$\frac{L}{v^2} = \frac{M}{2F},$$ the ratio of $L$ and $v^2$ is constant, as you noted.

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