Answer :
To determine the angular displacement of a figure skater who starts from rest and accelerates with a constant acceleration of 100 °/s² over a time period of 1 second, we can use the kinematic equation for rotational motion:
[tex]\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \][/tex]
where:
- [tex]\(\theta\)[/tex] is the angular displacement.
- [tex]\(\omega_0\)[/tex] is the initial angular velocity.
- [tex]\(\alpha\)[/tex] is the angular acceleration.
- [tex]\(t\)[/tex] is the time.
Given the problem parameters:
- The skater starts from rest, so the initial angular velocity [tex]\(\omega_0\)[/tex] is 0 °/s.
- The angular acceleration [tex]\(\alpha\)[/tex] is 100 °/s².
- The time [tex]\(t\)[/tex] is 1 second.
Plugging in these values:
[tex]\[ \theta = 0 \cdot 1 + \frac{1}{2} \cdot 100 \cdot 1^2 \][/tex]
[tex]\[ \theta = 0 + \frac{1}{2} \cdot 100 \cdot 1 \][/tex]
[tex]\[ \theta = \frac{100}{2} \][/tex]
[tex]\[ \theta = 50 \text{ degrees} \][/tex]
Thus, the angular displacement over 1 second is [tex]\(50^\circ\)[/tex].
Therefore, the correct answer is:
50°
[tex]\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \][/tex]
where:
- [tex]\(\theta\)[/tex] is the angular displacement.
- [tex]\(\omega_0\)[/tex] is the initial angular velocity.
- [tex]\(\alpha\)[/tex] is the angular acceleration.
- [tex]\(t\)[/tex] is the time.
Given the problem parameters:
- The skater starts from rest, so the initial angular velocity [tex]\(\omega_0\)[/tex] is 0 °/s.
- The angular acceleration [tex]\(\alpha\)[/tex] is 100 °/s².
- The time [tex]\(t\)[/tex] is 1 second.
Plugging in these values:
[tex]\[ \theta = 0 \cdot 1 + \frac{1}{2} \cdot 100 \cdot 1^2 \][/tex]
[tex]\[ \theta = 0 + \frac{1}{2} \cdot 100 \cdot 1 \][/tex]
[tex]\[ \theta = \frac{100}{2} \][/tex]
[tex]\[ \theta = 50 \text{ degrees} \][/tex]
Thus, the angular displacement over 1 second is [tex]\(50^\circ\)[/tex].
Therefore, the correct answer is:
50°
