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3 years ago

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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

1 answer:

Neko [114]3 years ago

5 0

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

$\alpha = 6261 \ rev/minutes^2$

b

$\theta = 613 \ revolutions$

Explanation:

From the question we are told that

The initial angular speed is $w_i = 1120 \ rev/minutes$

The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$ is $w_f = 2560 \ rev/minutes$

The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{2560 - 1120 }{0.23}$

=> $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$ is mathematically represented as

$\theta = \frac{1}{2} * (w_i + w_f)* t$

=> $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=> $\theta = 613 \ revolutions$

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