Solve the problem. -The grade resistance $\mathrm { F }$ of a car traveling up or down a hill is modeled by the equation $\mathrm { F } = \mathrm { W } \sin \theta$, where $W$ is the weight of the car and $\theta$ is the angle of the hill's grade $( \theta 0$ for uphill travel, $\theta

Question

Solve the problem.
-The grade resistance $\mathrm { F }$ of a car traveling up or down a hill is modeled by the equation $\mathrm { F } = \mathrm { W } \sin \theta$, where $W$ is the weight of the car and $\theta$ is the angle of the hill's grade $( \theta > 0$ for uphill travel, $\theta < 0$ for downhill travel). What is the grade resistance (to the nearest pound) of a $2500 - \mathrm { lb }$ car traveling downhill on a $6 ^ { \circ }$ grade $\left( \theta = - 6 ^ { \circ } \right)$ ?
A) $- 2506 \mathrm { lb }$
B) $261 \mathrm { lb }$
C) $2506 \mathrm { lb }$
D) $- 261 \mathrm { lb }$

Quizplus · Accepted Answer

The grade resistance $ F $ can be calculated using the given formula $ F = W \sin \theta $. For a $2500$-lb car traveling downhill on a $6^\circ$ grade ($\theta = -6^\circ$), we first convert $\theta$ to radians because the sine function typically uses radians. $ \theta = -6^\circ = -\frac{6\pi}{180} $ radians. Then, $ F = 2500 \sin(-\frac{6\pi}{180}) $. Using a calculator, $ \sin(-\frac{6\pi}{180}) \approx -0.1045 $, so $ F \approx 2500 \times -0.1045 \approx -261 $ lb, to the nearest pound.

Solve the Problem F\mathrm { F }F Of a Car Traveling Up or Down a Hill Is

Solve the Problem $\mathrm { F }$ Of a Car Traveling Up or Down a Hill Is