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Mathematics 23 Online
OpenStudy (anonymous):

Please help me understand what to do! When a car slams on its brakes and starts to skid, it leaves skid marks on the road. Investigators use these marks to determine how fast a car was going when it started to brake. The car's speed, v, is related to the skid mark length, d, by the equation v = sqrt(32 f d), where f is the coefficient of friction determined by the material the car was skidding on. Wet asphalt has a coefficient of friction of 0.5. If a car was driving 75 feet per second and slams on its brakes, how long of a skid mark will it leave? The skid marks will be ? feet long

OpenStudy (cwrw238):

plug in values into the formula: 75 = sqrt(32*0.5*d) 75 = sqrt(16d) solve for d

OpenStudy (cwrw238):

can you find d from this?

OpenStudy (cwrw238):

as you have a square root in the formula first square both side 75^2 = 16d 16d = 5625 d = 5625 / 16

OpenStudy (anonymous):

ouch

OpenStudy (unklerhaukus):

alternately; solve the equation for d first, \[v = \sqrt{32 f d}\]\[\frac{v}{\sqrt{32f}} = \sqrt{ d}\]\[{\frac{v^2}{{32f}}}=d\]then plug in the numbers for velocity and drag factor \[v=75[\text{ft/s}]\qquad f=0.5[\text{ft/s}^2]\] \[d={\frac{v^2}{{32f}}}\Longrightarrow{\frac{\left(75[\text{ft/s}]\right)^2}{{32\times0.5[\text{ft/s}^2]}}}\] \[\qquad\qquad =\frac{75^2}{32\times0.5}{\left[\frac{\text {ft}^2/\text{s}^2}{\text{ft/s}^2}\right]}\]\[\qquad d=\frac{75^2}{32\times0.5}{\left[\text {ft}\right]}\]

OpenStudy (anonymous):

@UnkleRhaukus So my is d= 75^2/32x0.5 ?

OpenStudy (anonymous):

answer*^

OpenStudy (anonymous):

I still don't get it ;( or how to get the answer?! I'm so lost!

OpenStudy (anonymous):

This is the hardest question ever!!!!!

OpenStudy (anonymous):

So is my answer 351.6 ?

ganeshie8 (ganeshie8):

\( \huge \color{green}{\checkmark} \)

OpenStudy (anonymous):

WOOOOHOOOO THANK YOU EVERYONE!

OpenStudy (unklerhaukus):

\[d=\frac{75^2}{32\times0.5}{\left[\text {ft}\right]}\]\[=\frac{(3\times5^2)^2}{2^5\times2^{-1}}{\left[\text {ft}\right]}\]\[=\frac{3^2\times5^4}{2^{5-1}}{\left[\text {ft}\right]}\]\[=\frac{3^2\times5^4}{2^{4}}\left[\text {ft}\right]\]\[=\frac{9\times625}{16}\left[\text {ft}\right]\]\[=351.5625\left[\text {ft}\right]\]

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