Question

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

Answer 1

plug in values into the formula:

75 = sqrt(32*0.5*d)
75 = sqrt(16d)

solve for d

Answer 2

can you find d from this?

Answer 3

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

Answer 4

ouch

Answer 5

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
\[v=75[\text{ft/s}]\qquad f=0.5[\text{ft/s}^2]\]
\[d={\frac{v^2}{{32f}}}\Longrightarrow{\frac{\left(75[\text{ft/s}]\right)^2}{{32\times0.5[\text{ft/s}^2]}}}\]
\[\qquad\qquad =\frac{75^2}{32\times0.5}{\left[\frac{\text {ft}^2/\text{s}^2}{\text{ft/s}^2}\right]}\]\[\qquad d=\frac{75^2}{32\times0.5}{\left[\text {ft}\right]}\]

Answer 6

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

Answer 7

answer*^

Answer 8

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

Answer 9

This is the hardest question ever!!!!!

Answer 10

So is my answer 351.6 ?

Answer 11

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

Answer 12

WOOOOHOOOO THANK YOU EVERYONE!

Answer 13

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