Answer:
The tub turns 37.520 revolutions during the 25-second interval.
Explanation:
The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:
[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)
Then, we expand the previous expression by kinematic equations for uniform accelerated motion:
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)
Where:
[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.
[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.
And each acceleration is determined by the following formulas:
Acceleration
[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)
Deceleration
[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)
Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.
If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:
[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]
[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]
[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]
[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]
[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]
[tex]\Delta n = 37.520\,rev[/tex]
The tub turns 37.520 revolutions during the 25-second interval.
