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Sagot :
To determine the acceleration of a 1520 kg car rolling down a frictionless hill with an inclination of [tex]\(14.4^{\circ}\)[/tex], we need to understand how gravity causes the car to accelerate down the hill.
### Step-by-Step Solution
1. Given Data:
- Mass of the car, [tex]\( m = 1520 \, \text{kg} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Angle of the incline, [tex]\( \theta = 14.4^{\circ} \)[/tex]
2. Convert the Angle to Radians:
- First convert the angle from degrees to radians.
[tex]\[ \theta_{\text{radians}} = \theta \times \frac{\pi}{180} \][/tex]
- For [tex]\( \theta = 14.4^{\circ} \)[/tex], we find:
[tex]\[ \theta_{\text{radians}} = 14.4 \times \frac{\pi}{180} \approx 0.2513 \, \text{radians} \][/tex]
3. Calculate the Component of Gravitational Force Along the Incline:
- On an inclined plane, the component of gravitational force causing acceleration along the hill is [tex]\( g \sin(\theta) \)[/tex].
- Using [tex]\(\theta = 0.2513 \, \text{radians}\)[/tex]:
[tex]\[ a = g \sin(\theta) \][/tex]
4. Calculate the Acceleration:
- Substituting the values into the equation:
[tex]\[ a = 9.8 \times \sin(0.2513) \approx 9.8 \times 0.2487 \approx 2.437 \, \text{m/s}^2 \][/tex]
Thus, the acceleration of the car down the frictionless hill is:
[tex]\[ a \approx 2.437 \, \text{m/s}^2 \][/tex]
### Step-by-Step Solution
1. Given Data:
- Mass of the car, [tex]\( m = 1520 \, \text{kg} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Angle of the incline, [tex]\( \theta = 14.4^{\circ} \)[/tex]
2. Convert the Angle to Radians:
- First convert the angle from degrees to radians.
[tex]\[ \theta_{\text{radians}} = \theta \times \frac{\pi}{180} \][/tex]
- For [tex]\( \theta = 14.4^{\circ} \)[/tex], we find:
[tex]\[ \theta_{\text{radians}} = 14.4 \times \frac{\pi}{180} \approx 0.2513 \, \text{radians} \][/tex]
3. Calculate the Component of Gravitational Force Along the Incline:
- On an inclined plane, the component of gravitational force causing acceleration along the hill is [tex]\( g \sin(\theta) \)[/tex].
- Using [tex]\(\theta = 0.2513 \, \text{radians}\)[/tex]:
[tex]\[ a = g \sin(\theta) \][/tex]
4. Calculate the Acceleration:
- Substituting the values into the equation:
[tex]\[ a = 9.8 \times \sin(0.2513) \approx 9.8 \times 0.2487 \approx 2.437 \, \text{m/s}^2 \][/tex]
Thus, the acceleration of the car down the frictionless hill is:
[tex]\[ a \approx 2.437 \, \text{m/s}^2 \][/tex]
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