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Sagot :
To model the torque [tex]\(\tau\)[/tex], we can represent it as a trigonometric function. Given the nature of the torque mentioned in the problem, a sine function is most suitable because it naturally oscillates between maximum and minimum values.
Here's how we derive the formula step-by-step:
1. Determine the amplitude [tex]\(A\)[/tex]:
- The amplitude of the sine function is half the difference between the maximum and minimum torque values.
- Given the maximum torque [tex]\(\tau_{\text{max}} = 0.01 \, \text{Nm}\)[/tex] and the minimum torque [tex]\(\tau_{\text{min}} = -0.01 \, \text{Nm}\)[/tex], we find:
[tex]\[ A = \frac{\tau_{\text{max}} - \tau_{\text{min}}}{2} = \frac{0.01 - (-0.01)}{2} = \frac{0.02}{2} = 0.01 \, \text{Nm} \][/tex]
2. Determine the angular frequency [tex]\(\omega\)[/tex]:
- The angular frequency is related to the period of the function. The period [tex]\(T\)[/tex] is the distance over which the function completes one full cycle, which is given as 1.2 meters.
- The angular frequency [tex]\(\omega\)[/tex] is given by:
[tex]\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{1.2 \, \text{m}} \approx 5.235987755982989 \, \text{radians per meter} \][/tex]
3. Determine the phase shift [tex]\(\phi\)[/tex]:
- The phase shift is the horizontal shift of the function to the right. The function reaches its maximum torque for the first time at 0.3 meters, indicating that the phase shift [tex]\(\phi\)[/tex] is 0.3 meters.
- Therefore, the phase shift [tex]\(\phi \approx 0.3 \, \text{meters}\)[/tex]
4. Form the torque function [tex]\(\tau(d)\)[/tex]:
- The general formula for the sine function considering amplitude [tex]\(A\)[/tex], angular frequency [tex]\(\omega\)[/tex], and phase shift [tex]\(\phi\)[/tex] is:
[tex]\[ \tau(d) = A \sin(\omega d - \phi) \][/tex]
- Substituting [tex]\(A\)[/tex], [tex]\(\omega\)[/tex], and [tex]\(\phi\)[/tex] we get:
[tex]\[ \tau(d) = 0.01 \sin(5.235987755982989 \, d - 0.3) \][/tex]
Now, let's find the torque when KC has ridden [tex]\(d = 4\)[/tex] meters into the race:
[tex]\[ \tau(4) = 0.01 \sin(5.235987755982989 \times 4 - 0.3) \][/tex]
Calculating the sine term:
[tex]\[ \tau(4) \approx 0.01 \sin(5.235987755982989 \times 4 - 0.3) \approx 0.0098 \, \text{Nm} \][/tex]
So, the torque when KC has ridden 4 meters is approximately 0.0098 Nm.
In conclusion, the formula for the torque [tex]\(\tau\)[/tex] applied by the gum on the tire as a function of the distance [tex]\(d\)[/tex] is:
[tex]\[ \tau(d) = 0.01 \sin(5.235987755982989 \, d - 0.3) \][/tex]
The torque when KC has ridden 4 meters into the race is approximately 0.0098 Nm.
Here's how we derive the formula step-by-step:
1. Determine the amplitude [tex]\(A\)[/tex]:
- The amplitude of the sine function is half the difference between the maximum and minimum torque values.
- Given the maximum torque [tex]\(\tau_{\text{max}} = 0.01 \, \text{Nm}\)[/tex] and the minimum torque [tex]\(\tau_{\text{min}} = -0.01 \, \text{Nm}\)[/tex], we find:
[tex]\[ A = \frac{\tau_{\text{max}} - \tau_{\text{min}}}{2} = \frac{0.01 - (-0.01)}{2} = \frac{0.02}{2} = 0.01 \, \text{Nm} \][/tex]
2. Determine the angular frequency [tex]\(\omega\)[/tex]:
- The angular frequency is related to the period of the function. The period [tex]\(T\)[/tex] is the distance over which the function completes one full cycle, which is given as 1.2 meters.
- The angular frequency [tex]\(\omega\)[/tex] is given by:
[tex]\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{1.2 \, \text{m}} \approx 5.235987755982989 \, \text{radians per meter} \][/tex]
3. Determine the phase shift [tex]\(\phi\)[/tex]:
- The phase shift is the horizontal shift of the function to the right. The function reaches its maximum torque for the first time at 0.3 meters, indicating that the phase shift [tex]\(\phi\)[/tex] is 0.3 meters.
- Therefore, the phase shift [tex]\(\phi \approx 0.3 \, \text{meters}\)[/tex]
4. Form the torque function [tex]\(\tau(d)\)[/tex]:
- The general formula for the sine function considering amplitude [tex]\(A\)[/tex], angular frequency [tex]\(\omega\)[/tex], and phase shift [tex]\(\phi\)[/tex] is:
[tex]\[ \tau(d) = A \sin(\omega d - \phi) \][/tex]
- Substituting [tex]\(A\)[/tex], [tex]\(\omega\)[/tex], and [tex]\(\phi\)[/tex] we get:
[tex]\[ \tau(d) = 0.01 \sin(5.235987755982989 \, d - 0.3) \][/tex]
Now, let's find the torque when KC has ridden [tex]\(d = 4\)[/tex] meters into the race:
[tex]\[ \tau(4) = 0.01 \sin(5.235987755982989 \times 4 - 0.3) \][/tex]
Calculating the sine term:
[tex]\[ \tau(4) \approx 0.01 \sin(5.235987755982989 \times 4 - 0.3) \approx 0.0098 \, \text{Nm} \][/tex]
So, the torque when KC has ridden 4 meters is approximately 0.0098 Nm.
In conclusion, the formula for the torque [tex]\(\tau\)[/tex] applied by the gum on the tire as a function of the distance [tex]\(d\)[/tex] is:
[tex]\[ \tau(d) = 0.01 \sin(5.235987755982989 \, d - 0.3) \][/tex]
The torque when KC has ridden 4 meters into the race is approximately 0.0098 Nm.
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