The table shows data for a car stopping on a wet road. Stopping distance formula Determine the value of f, rounded to the nearest hundredth.

We have that the value of f, rounded to the nearest hundredth.

f_1=0.02

f_2=0.009

f_3=0.009

From the question we are told that

d(v)= \frac{2.15v^2}{64.4f}

Generally the equation for the f is mathematically given as

f= \frac{2.15v^2}{64.4*d(v)}

Therefore

For v=20 and f=38

f= \frac{2.15(20)^2}{64.4*(38)(20)}

f=0.02

For v=30 and f=86

f=\frac{2.15(30)^2}{64.4*(86)(30)}

f=0.01

For v=40 and f=153

f=\frac{2.15(40)^2}{64.4*(153)(40)}

f=0.009

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0.35

the next one is d(v) = 2.15v^2 / 22.54

Step-by-step explanation:

i just did this assignment. also thanks to the answer above me :)

the answer is 31.27mph

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.