on an expressway, the recommended safe distance between cars in feet is given by 0.019v^2+v+14, where v is the speed of the car in miles per hour. Find the safe distance when the following takes place( round answer to three decimal places). v=35 mph ____ft.? and v=55mph _____ft.?
well, just plug in your numbers:
f(35) = 0.019*35^2 + 35 + 14 = 72.275
and so on with other values
To find the safe distance between cars at a given speed, we will substitute the given speeds into the equation for the recommended safe distance.
First, let's find the safe distance when v = 35 mph.
Step 1: Substitute the value of v into the equation for safe distance:
Distance = 0.019v^2 + v + 14
Distance = 0.019(35)^2 + 35 + 14
Step 2: Simplify the equation:
Distance = 0.019(1225) + 35 + 14
Distance = 23.275 + 35 + 14
Distance ≈ 72.275 feet (rounded to three decimal places)
Therefore, when v = 35 mph, the safe distance between cars is approximately 72.275 feet.
Next, let's find the safe distance when v = 55 mph.
Step 1: Substitute the value of v into the equation for safe distance:
Distance = 0.019v^2 + v + 14
Distance = 0.019(55)^2 + 55 + 14
Step 2: Simplify the equation:
Distance = 0.019(3025) + 55 + 14
Distance = 57.475 + 55 + 14
Distance ≈ 126.475 feet (rounded to three decimal places)
Therefore, when v = 55 mph, the safe distance between cars is approximately 126.475 feet.