Answer:
a. yes
Explanation:
The initial speed of the circular saw is:
[tex]\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}[/tex]
[tex]\dot n_{o} = 100\,\frac{rev}{s}[/tex]
Deceleration rate needed to stop the circular saw is:
[tex]\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}[/tex]
[tex]\ddot n = - 50\,\frac{rev}{s^{2}}[/tex]
The number of turns associated with such deceleration rate is:
[tex]\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}[/tex]
[tex]\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}[/tex]
[tex]\Delta n = 100\,rev[/tex]
Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.
