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andrey2020 [161]
3 years ago
9

Assume that the operating cost of a certain truck (excluding driver's wages) is 12+x/6 cents per mile when the truck travels at

x mi/hr. If the driver earns $6 per hour, what is the most economical speed to operate the truck on a 400 mi turnpike where the minimum speed is 40 mph and the maximum speed is 70 mph?
Physics
1 answer:
podryga [215]3 years ago
8 0

Answer:

x = 60 mph

Explanation:

Given that the operating cost is

c = 12 + \frac{x}{6} cents per mile

total miles covered is given as

d = 400 miles

so total cost of drive is given as

C = (12 + \frac{x}{6})(4) $

time taken by the truck to move the distance is given as

t = \frac{400}{x}

So total earnings of the driver is given as

E = \frac{400}{x} \times 6 $

now total profit of the driver is given as

P = \frac{2400}{x} - (48 + \frac{2x}{3}) $

to maximize the profit we have

\frac{dP}{dx} = 0

-\frac{2400}{x^2} + \frac{2}{3} = 0

so we have

x = 60 mph

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Cylinder:

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Inner radius of the cylinder (R_i) = 0.2 m

Outer radius of the cylinder (R_o) = 0.3 m

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Now

The rotational Kinetic energy.

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Plugging the values.

⇒  (KE)_r=\frac{0.566\times (94.876)^2}{2}

⇒ (KE)_r=2547.41 Joules

Then

The kinetic energy of the rotational system is 2547.41 J.

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