Question

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 2500-lb car traveling downhill on a ( 6^circ ) grade (( \theta = -6^circ ))?
a. 2506 lb
b. -261 lb
c. -2506 lb
d. 261 lb

Explanation:

Step1: Identify given values

We know that \( W = 2500 \) lb and \( \theta=- 6^{\circ} \), and the formula for grade resistance is \( F = W\sin\theta \).

Step2: Substitute values into formula

Substitute \( W = 2500 \) and \( \theta=-6^{\circ} \) into the formula: \( F=2500\times\sin(- 6^{\circ}) \).
We know that \( \sin(-\alpha)=-\sin\alpha \), so \( \sin(-6^{\circ})=-\sin(6^{\circ}) \).
First, calculate \( \sin(6^{\circ})\approx0.1045 \).
Then \( F = 2500\times(- 0.1045)=- 261.25\approx - 261 \) lb.

Answer:

B. -261 lb