Complete Question
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.
(a) What is its angular acceleration in revolutions per minute-squared
(b) How many revolutions does the engine make during this 20 s interval?
rev
Answer:
a
[tex]\alpha = 6261 \ rev/minutes^2[/tex]
b
[tex]\theta = 613 \ revolutions[/tex]
Explanation:
From the question we are told that
The initial angular speed is [tex]w_i = 1120 \ rev/minutes[/tex]
The angular speed after [tex]t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes[/tex] is [tex]w_f = 2560 \ rev/minutes[/tex]
The time for revolution considered is [tex]t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes[/tex]
Generally the angular acceleration is mathematically represented as
[tex]\alpha = \frac{w_f - w_i }{t}[/tex]
=> [tex]\alpha = \frac{2560 - 1120 }{0.23}[/tex]
=> [tex]\alpha = 6261 \ rev/minutes^2[/tex]
Generally the number of revolution made is [tex]t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes[/tex] is mathematically represented as
[tex]\theta = \frac{1}{2} * (w_i + w_f)* t[/tex]
=> [tex]\theta = \frac{1}{2} * (1120+ 2560 )* 0.333[/tex]
=> [tex]\theta = 613 \ revolutions[/tex]
