Solve the problem. -The grade resistance $F$ of a car traveling up or down a hill is modeled by the equation $F = W \sin \theta$, 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 $F$ of a car traveling up or down a hill is modeled by the equation $F = W \sin \theta$, 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 $1000 - \mathrm { lb }$ car traveling uphill on a $1 ^ { \circ } \operatorname { grade } \left( \theta = 1 ^ { \circ } \right)$ ?
A) $- 17 \mathrm { lb }$
B) $1001 \mathrm { lb }$
C) $- 1001 \mathrm { lb }$
D) $17 \mathrm { lb }$

Quizplus · Accepted Answer

To find the grade resistance, substitute $W = 1000 \, \mathrm{lb}$ and $\theta = 1^\circ$ into the equation $F = W \sin \theta$. First, convert $\theta$ to radians because the sine function in most calculators requires radians. $1^\circ = \frac{\pi}{180}$ radians. Then, $F = 1000 \sin\left(\frac{\pi}{180}\right)$. Calculating this gives a value close to 17 lb, thus the grade resistance is approximately 17 lb.

Solve the Problem FFF Of a Car Traveling Up or Down a Hill Is

Solve the Problem $F$ Of a Car Traveling Up or Down a Hill Is