Question

ASearch PHYS 110 Course Home Problem 8.22 6of15 (?) PartA Constants The tires of a car make 94 revolutions as the car reduces its speed uniformly from 94.0 km/h to 65.0 km/h. The tires have a diamoter of 0.86 m. What was the angular aoceleration of the tires? Express your answer using two significant figures. Correct Part B ifthe car continues to decelerate at this rato, how much more time is required fr it to stop? Express your answer to two significant figures and include the appropriate units Value Submit Incorrect: Try Again Part c

Answer 1

a)
Total angular displacement, $\theta$ = 94 rev
= 94 x 2 $\pi$ rad
= 590.62 rad

Initial speed, Vi = 94 km/h = 26.11 m/s
Initial angular velocity, $\omega$ i = Vi/R
= 26.11/0.43
= 60.72 rad/s

Final speed, Vf = 65 km/h = 18.06 m/s
Final angular velocity, $\omega$ f = Vf/R
= 18.06/0.43
= 41.99 rad/s

Using the relation, ( $\omega$ f)² - ( $\omega$ i)² = 2 $\alpha$ $\theta$ ,
Angular acceleration, $\alpha$ = [( $\omega$ f)² - ( $\omega$ i)²] / 2 $\theta$
= [(41.99)² - ( 60.72)²] / [2 x 590.62]
= - 1.63 rad/s²

b)
Initial angular velocity, $\omega$ i = 41.99 rad/s
Final angular velocity, $\omega$ f = 0
Acceleration, $\alpha$ = - 1.63 rad/s²
Using the formula, $\omega$ f = $\omega$ i + $\alpha$ t,
t = ( $\omega$ f - $\omega$ i) / $\alpha$
= (0 - 41.99) / (-1.63)
= 25.78 s

c)
Using the relation, ( $\omega$ f)² - ( $\omega$ i)² = 2 $\alpha$ $\theta$ ,
Where $\omega$ f = 0, $\omega$ i = 41.99 rad/s, angular acceleration, $\alpha$ = - 1.63 rad/s²,
$\theta$ = [( $\omega$ f)² - ( $\omega$ i)²] / 2 $\alpha$
= [0 - 41.99²] / [2 x - 1.63]
= 541.18 rad
= 86.13 rev
One revolution = 2 $\pi$ r meters
Total distance = 86.13 x 2 $\pi$ x 0.43
= 232.71 m