Answer:
[tex]a_r=389\ rev.s^{-2}[/tex]
[tex]n=58350 rev[/tex]
Explanation:
Given:
time of constant deceleration, [tex]t=10\ s[/tex]
A.
initial angular speed, [tex]N_i=3890\ rpm\ [/tex]
Using equation of motion:
[tex]N_f=N_i+a_r.t[/tex]
[tex]0=3890+a_r\times 10[/tex]
[tex]a_r=389\ rev.s^{-2}[/tex]
B.
Using eq. of motion for no. of revolutions, we have:
[tex]n=N_i.t+\frac{1}{2} a_r.t^2[/tex]
[tex]n=3890\times 10+0.5\times 389\times 100[/tex]
[tex]n=58350\ rev[/tex]