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Sagot :
To find the final velocity of the truck as it slides down the frictionless hill, we'll follow these steps:
1. Identify the given values:
- Mass of the truck, [tex]\( m = 3220 \, \text{kg} \)[/tex] (not directly needed for this problem since we are dealing with a frictionless scenario).
- Initial velocity of the truck, [tex]\( v_i = 4.71 \, \text{m/s} \)[/tex].
- Length of the hill, [tex]\( d = 22.0 \, \text{m} \)[/tex].
- Angle of the hill, [tex]\( \theta = 8.30^\circ \)[/tex].
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex].
2. Convert the hill angle from degrees to radians for the subsequent trigonometric calculations:
[tex]\[ \theta_{radians} = 0.14486232791552936 \, \text{radians} \][/tex]
3. Calculate the component of the gravitational acceleration along the hill:
The acceleration component parallel to the hill is given by:
[tex]\[ g_{\parallel} = g \sin(\theta) = 1.4161343318195472 \, \text{m/s}^2 \][/tex]
4. Apply the kinematic equation to find the final velocity:
The kinematic equation for an object initially moving with velocity [tex]\( v_i \)[/tex] and then accelerating over a distance [tex]\( d \)[/tex] is:
[tex]\[ v_f^2 = v_i^2 + 2 a d \][/tex]
Here, [tex]\( a = g_{\parallel} \)[/tex], so:
[tex]\[ v_f^2 = v_i^2 + 2 \cdot g_{\parallel} \cdot d \][/tex]
Plugging in the known values:
[tex]\[ v_f^2 = (4.71)^2 + 2 \cdot 1.4161343318195472 \cdot 22.0 \][/tex]
[tex]\[ v_f^2 = 22.1841 + 62.30991059634 = 84.49401060006008 \][/tex]
5. Take the square root of both sides to find the final velocity [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \sqrt{84.49401060006008} \approx 9.192062369243372 \, \text{m/s} \][/tex]
Thus, the final velocity [tex]\( v_f \)[/tex] of the truck is approximately:
[tex]\[ v_f = 9.192 \, \text{m/s} (rounded to three significant digits) \][/tex]
1. Identify the given values:
- Mass of the truck, [tex]\( m = 3220 \, \text{kg} \)[/tex] (not directly needed for this problem since we are dealing with a frictionless scenario).
- Initial velocity of the truck, [tex]\( v_i = 4.71 \, \text{m/s} \)[/tex].
- Length of the hill, [tex]\( d = 22.0 \, \text{m} \)[/tex].
- Angle of the hill, [tex]\( \theta = 8.30^\circ \)[/tex].
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex].
2. Convert the hill angle from degrees to radians for the subsequent trigonometric calculations:
[tex]\[ \theta_{radians} = 0.14486232791552936 \, \text{radians} \][/tex]
3. Calculate the component of the gravitational acceleration along the hill:
The acceleration component parallel to the hill is given by:
[tex]\[ g_{\parallel} = g \sin(\theta) = 1.4161343318195472 \, \text{m/s}^2 \][/tex]
4. Apply the kinematic equation to find the final velocity:
The kinematic equation for an object initially moving with velocity [tex]\( v_i \)[/tex] and then accelerating over a distance [tex]\( d \)[/tex] is:
[tex]\[ v_f^2 = v_i^2 + 2 a d \][/tex]
Here, [tex]\( a = g_{\parallel} \)[/tex], so:
[tex]\[ v_f^2 = v_i^2 + 2 \cdot g_{\parallel} \cdot d \][/tex]
Plugging in the known values:
[tex]\[ v_f^2 = (4.71)^2 + 2 \cdot 1.4161343318195472 \cdot 22.0 \][/tex]
[tex]\[ v_f^2 = 22.1841 + 62.30991059634 = 84.49401060006008 \][/tex]
5. Take the square root of both sides to find the final velocity [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \sqrt{84.49401060006008} \approx 9.192062369243372 \, \text{m/s} \][/tex]
Thus, the final velocity [tex]\( v_f \)[/tex] of the truck is approximately:
[tex]\[ v_f = 9.192 \, \text{m/s} (rounded to three significant digits) \][/tex]
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