Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

A car in motion starts to coast up a frictionless hill that is inclined at [tex][tex]$14.0^{\circ}$[/tex][/tex].

What is the acceleration of the car?

[tex]a = [?] \, \text{m/s}^2[/tex]


Sagot :

To determine the acceleration of a car coasting up a frictionless hill that is inclined at [tex]\(14.0^{\circ}\)[/tex], we need to consider the forces acting on the car due to gravity. Specifically, we will use the component of gravitational force that acts along the direction of the incline to find the acceleration.

Here is the detailed step-by-step solution:

1. Identify the angle of the incline: The hill is inclined at [tex]\(14.0^{\circ}\)[/tex].

2. Convert the angle from degrees to radians: Calculations in physical formulas involving trigonometric functions often require angles in radians. The conversion from degrees to radians can be done using the conversion factor [tex]\(\pi \text{ radians} = 180^\circ\)[/tex]. Thus,
[tex]\[ \theta = 14.0^\circ \times \frac{\pi \text{ radians}}{180^\circ} = 0.24434609527920614 \text{ radians} \][/tex]

3. Understand the gravitational component causing acceleration: On an inclined plane, the component of the gravitational force that causes acceleration down the incline is given by [tex]\( g \sin(\theta) \)[/tex], where [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\(9.8 \, \text{m/s}^2\)[/tex]) and [tex]\( \theta \)[/tex] is the angle of inclination.

4. Calculate the acceleration: Using the angle in radians and the gravitational constant,
[tex]\[ a = g \sin(\theta) = 9.8 \, \text{m/s}^2 \times \sin(0.24434609527920614) \][/tex]

5. Result of the acceleration calculation:
[tex]\[ a = 2.370834576876744 \, \text{m/s}^2 \][/tex]

Hence, the acceleration of the car as it coasts up the frictionless hill inclined at [tex]\(14.0^\circ\)[/tex] is approximately:
[tex]\[ a \approx 2.37 \, \text{m/s}^2 \][/tex]