Question

A car is traveling along a road, and its engine is turning over with an angular velocity of +188 rad/s. The driver steps on the accelerator, and in a time of 15.6 s the angular velocity increases to +293 rad/s. (a) What would have been the angular displacement of the engine if its angular velocity had remained constant at the initial value of +188 rad/s during the entire 15.6-s interval? (b) What would have been the angular displacement if the angular velocity had been equal to its final value of +293 rad/s during the entire 15.6-s interval? (c) Determine the actual value of the angular displacement during the 15.6-s interval.

Answer 1

Answer:

3751.80514 radians

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=188\times 15.6+\dfrac{1}{2}\times 0\times 15.6^2\\\Rightarrow \theta=2932.8\ rad$

The angular displacement would be 2932.8 rad

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=293\times 15.6+\dfrac{1}{2}\times 0\times 15.6^2\\\Rightarrow \theta=4570.8\ rad$

The angular displacement would be 4570.8 rad

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{293-188}{15.6}\\\Rightarrow \alpha=6.73076\ rad/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\dfrac{293^2-188^2}{2\times 6.73076}\\\Rightarrow \theta=3751.80514\ rad$

Actual value of the angular displacement is 3751.80514 radians