Question

the function ( v(r) = sqrt{2.6r} ) models the maximum safe speed, ( v ), in miles per hour, at which a car can travel on a curved road with radius of curvature ( r ), in feet. a highway crew measures the radius of curvature at an exit ramp on a highway as 650 feet. what is the maximum safe speed?
for this problem, round your answer down to the nearest whole number. (think: why is this type of rounding appropriate for this scenario?)
max safe speed = (square) mph

Explanation:

Step1: Identify the formula and input

The formula for maximum safe speed is \( v(r) = \sqrt{2.6r} \), and the radius \( r = 650 \) feet.

Step2: Substitute \( r \) into the formula

Substitute \( r = 650 \) into \( v(r) \): \( v(650)=\sqrt{2.6\times650} \)

Step3: Calculate the product inside the square root

First, calculate \( 2.6\times650 = 1690 \)

Step4: Take the square root

Then, \( \sqrt{1690}\approx41.11 \)

Step5: Round down to the nearest whole number

Since we need to round down, the value is \( 41 \)

Answer:

41