The distance it takes a truck to stop can be modeled by the function

Determine the value off, rounde

hundredth.

2

DONE

2.150 ?

d(u)=-

64.4f

d = stopping distance in feet

y = initial velocity in miles per hour

f= a constant related to friction

When the truck's initial velocity on dry pavement is

40 mph, its stopping distance is 138 ft.

Your question is not well presented (Refer below for correct question)

The distance it takes a truck to stop can be modeled by the function;

d(u) = 2.15u²/64.4f

Where

d = stopping distance in feet

u = initial velocity in miles per hour

f = a constant related to friction

When the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft.

Determine the value of f.

The friction constant is approximately 0.3871

Step-by-step explanation:

Given

Stopping distance, d(u) = 138ft

Initial Velocity, u = 40mph

Required

Friction constant, f.

To get the value of f, we need to simply substitute 40 for u and 138 for d(u) in the above expression.

In other words;

d(u) = 2.15u²/64.4f becomes

138 = 2.15 * 40²/64.4f

138 = 2.15 * 1600/64.4f

138 = 3440/64.4f

Multiply both sides by 64.4f

138 * 64.4f = 64.4f * 3440/64.4f

8887.2f = 3440

Divide both sides by 8887.2

8887.2f/8887.2 = 3440/8887.2

f = 3440/8887.2

f = 0.387073544

f = 0.3871 (Approximated)

Hence, the friction constant is approximately 0.3871

6.5

step-by-step explanation:

you said it was in the question

~courtney

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64