(Modeling) Grade Resistance Solve each problem. See Example 3. A 3000-lb car traveling uphill has a grade resistance of 150 lb. Find the angle of the grade to the nearest tenth of a degree.

Verified step by step guidance

1. Understand that the grade resistance is the component of the car's weight acting parallel to the slope. This force can be expressed as \(W \sin(\theta)\), where \(W\) is the weight of the car and \(\theta\) is the angle of the grade.
2. Set up the equation relating the grade resistance to the weight and the sine of the angle: \(150 = 3000 \sin(\theta)\).
3. Isolate \(\sin(\theta)\) by dividing both sides of the equation by 3000: \(\sin(\theta) = \frac{150}{3000}\).
4. Use the inverse sine function to find the angle \(\theta\): \(\theta = \sin^{-1}\left(\frac{150}{3000}\right)\).
5. Calculate the angle \(\theta\) using a calculator and round the result to the nearest tenth of a degree.

Key Concepts

Grade Resistance and Its Relation to Weight

Grade resistance is the component of a vehicle's weight acting parallel to an inclined surface, opposing motion uphill. It depends on the weight of the vehicle and the angle of the incline, representing the force needed to overcome gravity along the slope.

Trigonometric Relationship Between Forces and Angles

The grade resistance corresponds to the component of the weight vector along the incline, which can be expressed as weight times the sine of the incline angle. Using the sine function allows us to relate the known forces to the unknown angle.

Inverse Sine Function to Find the Angle

To find the angle of the grade, we use the inverse sine (arcsin) function on the ratio of grade resistance to weight. This operation retrieves the angle whose sine equals the given ratio, enabling calculation of the incline angle in degrees.