Question

more than 30°. find an interval for how far the rider travels horizontally.

1. speed enforcement a police department has set up a speed enforcement zone on a straight length of highway. a patrol car is parked parallel to the zone, 200 feet from one end and 150 feet from the other end (see figure).

(a) find the length / of the zone and the measures of the angles a and b (in degrees).
(b) find the minimum amount of time (in seconds) it takes for a vehicle to pass through the zone without exceeding the posted speed limit of 35 miles per hour.

Explanation:

Step1: Use Pythagorean theorem for length \( l \)

By Pythagorean theorem \( l=\sqrt{200^{2}+150^{2}}=\sqrt{40000 + 22500}=\sqrt{62500}=250\) feet.

Step2: Use trigonometric ratios for angles \( A \) and \( B \)

For angle \( A\), \(\tan A=\frac{150}{200}=0.75\), so \(A=\arctan(0.75)\approx36.87^{\circ}\)
For angle \( B\), \(\tan B=\frac{200}{150}=\frac{4}{3}\approx1.333\), so \(B=\arctan(\frac{4}{3})\approx53.13^{\circ}\)

Step3: Convert speed from miles per hour to feet per second

Since \(1\) mile \( = 5280\) feet and \(1\) hour \(=3600\) seconds. The speed \(v = 35\) miles per hour.
\(v=35\times\frac{5280}{3600}=\frac{35\times22}{15}=\frac{770}{15}=\frac{154}{3}\approx51.33\) feet per second.

Step4: Use time formula \(t=\frac{d}{v}\)

Given \(d = 250\) feet and \(v=\frac{154}{3}\) feet per second.
\(t=\frac{250}{\frac{154}{3}}=\frac{750}{154}\approx4.87\) seconds.

Answer:

(a) The length \(l = 250\) feet, \(A\approx36.87^{\circ}\), \(B\approx53.13^{\circ}\)
(b) The minimum time \(t\approx4.87\) seconds.