The number of revolutions the tire makes during this motion is 0.95 revolutions and the final angular speed of a tire in revolutions per second is 75.86 rad/sec.
Given :
A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. Assuming the diameter of a tire is 58.0 cm.
a) First determine the acceleration from the given data.
v = u + at
where v is the final velocity, u is the initial velocity. a is the acceleration, and t is the time.
Now, substitute the value of known terms in the above equation.
22 = a(9)
a = 2.44m/[tex]\rm s^2[/tex]
Now, determine the displacement by using the below equation.
[tex]\rm S = ut + \dfrac{1}{2}at^2[/tex]
where 'S' is the displacement, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time in seconds.
Now, substitute the values of known terms in the above equation.
[tex]\rm S = 0\times 9 +\dfrac{1}{2}\times 2.44 \times 9^2[/tex]
[tex]\rm S = 98.82\;m[/tex]
Now, use the below formula in order to determine the number of revolutions.
[tex]\rm S = r \times \theta[/tex]
[tex]\dfrac{98.82}{0.29}=\theta[/tex]
[tex]\theta = 341.44^\circ[/tex]
Remember one revolution is of 360 degrees so, in [tex]341.44^\circ[/tex] there is a total of:
[tex]=\dfrac{341.44}{360}[/tex]
= 0.95 revolotion.
b) The angular speed is given by the formula:
[tex]\omega = \dfrac{22}{0.29}[/tex]
[tex]\rm \omega = 75.86\;rad/sec[/tex]
For more information, refer to the link given below:
https://brainly.com/question/2635551
