Answer :
Answer:
The total linear acceleration is approximately 0.246 meters per square second.
Explanation:
The total linear acceleration ([tex]a[/tex]) consist in two components, radial ([tex]a_{r}[/tex]) and tangential ([tex]a_{t}[/tex]), in meters per square second:
[tex]a_{r} = \omega^{2}\cdot r[/tex] (1)
[tex]a_{t} = \alpha \cdot r[/tex] (2)
Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:
[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex] (3)
Where:
[tex]r[/tex] - Radius of the wheel, in meters.
[tex]\omega[/tex] - Angular speed, in radians per second.
[tex]\alpha[/tex] - Angular acceleration, in radians per square second.
Given that wheel accelerates uniformly, we use the following kinematic equation:
[tex]\omega = \omega_{o}+ \alpha\cdot t[/tex] (4)
Where:
[tex]\omega_{o}[/tex] - Initial angular speed, in radians per second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]r = 0.1\,m[/tex], [tex]\alpha = 2\,\frac{rad}{s^{2}}[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex] and [tex]t = 0.60\,s[/tex], then the total linear acceleration is:
[tex]\omega = \omega_{o}+ \alpha\cdot t[/tex]
[tex]\omega = 1.2\,\frac{rad}{s}[/tex]
[tex]a_{r} = \omega^{2}\cdot r[/tex]
[tex]a_{r} = 0.144\,\frac{m}{s^{2}}[/tex]
[tex]a_{t} = \alpha \cdot r[/tex]
[tex]a_{t} = 0.2\,\frac{m}{s^{2}}[/tex]
[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex]
[tex]a \approx 0.246\,\frac{m}{s^{2}}[/tex]
The total linear acceleration is approximately 0.246 meters per square second.