Question

Beginning drivers soon learn that bringing an automobile to a stop requires greater distances for higher speeds. A formula that approximates the stopping distance under normal driving conditions is
d = 1.1v + 0.055v2
where v = speed of the auto in miles per hour and d = distance in feet required to bring the auto to a stop.
(a) Find the stopping distance of an auto traveling 15 mph. (Round your answer to the nearest foot.)
ft

(b) Find the stopping distance of an auto traveling 30 mph. (Round your answer to the nearest foot.)
ft

Answer 1

The formula for the stopping distance in feet is

$d=1.1v+0.055v^2$

where $v$ = Speed of the car in miles per hour.

The stopping distance for the car traveling at 15 mph is

$d=1.1\times 15+0.055\times 15^2\\\Rightarrow d=28.875\approx 29\ \text{feet}$

The stopping distance for the auto traveling at 15 mph is $29\ \text{feet}$.

The stopping distance for the car traveling at 30 mph is

$d=1.1\times 30+0.055\times 30^2\\\Rightarrow d=82.5\approx 83\ \text{feet}$

The stopping distance for the auto traveling at 30 mph is $83\ \text{feet}$.

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