the table below shows the data for a car stopping on a wet road. what is the approximate stopping distance…

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?\ncar stopping distances\nv (mph) d (ft)\n15 17.9\n20 31.8\n50 198.7\nd(v)=\\frac{2.15v^{2}}{64.4f}\n41.7 ft\n49.7 ft\n97.4 ft\n115.3 ft

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?\ncar stopping distances\nv (mph) d (ft)\n15 17.9\n20 31.8\n50 198.7\nd(v)=\\frac{2.15v^{2}}{64.4f}\n41.7 ft\n49.7 ft\n97.4 ft\n115.3 ft

Answer

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