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any idea?
well you could recognize that the speed is twice as fast, and therefore the square of the speed is four times as large. Or if you prefer you can set it up as an equation in terms of the speed r, and a constant of proportionality \(\alpha\) \[d(r)=\alpha r^2\]Now solving for alpha\[d(50)=\alpha(50^2)=110\]\[\alpha=\frac{110}{50^2}\]And then simply plug 100mph into the d(r)
1100?
how did you get that?
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is it wrong?
I dunno. I can't see your work.
Ok thanks
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