Question 16, 2B.4.95
Part 1 of 2
An accepted relationship between stopping distance, $d$ in feet, and the speed of a car, $v$ in mph, is $d(v)=1.3 v+0.04 v$
(a) How many feet will it take a car traveling $40 \mathrm{mph}$ to stop on dry, level concrete?
(b) If an accident occurs 150 feet ahead, what is the maximum speed at which one can travel to avoid being involved
(a) It will take $\mathrm{ft}$ to stop the car that is traveling at $40 \mathrm{mph}$.
(Round up to the nearest foot as needed.)
Answer
Final Answer: It will take \(\boxed{116}\) feet to stop the car that is traveling at \(40 \mathrm{mph}\).
Step 1 ：The question is asking for the stopping distance of a car traveling at 40 mph. The formula for stopping distance is given as \(d(v)=1.3 v+0.04 v^2\). We can substitute \(v=40\) into this formula to find the stopping distance.
Step 2 ：Substitute \(v=40\) into the formula: \(d = 1.3(40) + 0.04(40^2)\)
Step 3 ：Solve the equation to find the stopping distance: \(d = 116.0\)
Step 4 ：Round the stopping distance to the nearest foot: \(d = 116\)
Step 5 ：Final Answer: It will take \(\boxed{116}\) feet to stop the car that is traveling at \(40 \mathrm{mph}\).
