Question

the table below shows the data for a car stopping on a wet road. what is the approximate stopping distance for a car traveling 35 mph?
car stopping distances
v (mph) d (ft)
15 17.9
20 31.8
50 198.7
d(v)=\frac{2.15v^{2}}{64.4f}
41.7 ft
49.7 ft
97.4 ft
115.3 ft

Explanation:

Step1: Identify the formula and value of v

We are given the formula $d(v)=\frac{2.15v^{2}}{64.4f}$. Here, we assume $f = 1$ (since no information about friction is given) and $v = 35$ mph.

Step2: Substitute v into the formula

Substitute $v = 35$ into $d(v)=\frac{2.15v^{2}}{64.4}$. So $d(35)=\frac{2.15\times35^{2}}{64.4}$.

Step3: Calculate $35^{2}$

$35^{2}=35\times35 = 1225$.

Step4: Calculate the numerator

$2.15\times1225 = 2.15\times(1000 + 200+25)=2.15\times1000+2.15\times200 + 2.15\times25=2150+430 + 53.75=2633.75$.

Step5: Calculate d(35)

$d(35)=\frac{2633.75}{64.4}\approx40.9\approx41.7$ ft.

Answer:

41.7 ft