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During a very quick stop, a car slows down at 5.40 m/s². Assume the tires rotate in conterclockwise direction. (a) What is the angular acceleration of its 0.275 m-radius tires, assuming they do not slip on the pavement? Pay attention to the sign. Enter a number rad/s2 (5 attempts remaining)Input 1 Status (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 83.5 rad/s? Enter a number rev (5 attempts remaining)Input 2 Status (c) How long in seconds does the car take to stop completely from the initital angular velocity in (b)? Enter a number s (5 attempts remaining)Input 3 Status (d) What distance in meters does the car travel in this time? Enter a number m (5 attempts remaining)Input 4 Status (e) What was the car’s initial velocity in m/s?

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skyluke89

a) The angular acceleration is [tex]-19.6 rad/s^2[/tex]

b) The tires make 28.3 revolutions before coming to rest

c) The car takes 4.26 s to stop

d) The distance travelled by the car is 49.0 m

e) The initial velocity of the car is 23.0 m/s

Explanation:

a)

The linear acceleration of the tires of the car is

[tex]a=-5.40 m/s^2[/tex]

where the negative sign is due to the fact that the car is slowing down.

The radius of the tires is

r = 0.275 m

We know that the angular acceleration is related to the linear acceleration by

[tex]a=\alpha r[/tex]

or

[tex]\alpha = \frac{a}{r}[/tex]

And substituting, we find

[tex]\alpha = \frac{-5.40}{0.275}=-19.6 rad/s^2[/tex]

b)

The angular displacement of the tire can be found by using the equivalen of the suvat equation for rotational motions:

[tex]\omega_f^2 - \omega_i^2 = 2 \alpha \theta[/tex]

where

[tex]\omega_0 = 83.5 rad/s[/tex] is the initial angular velocity

[tex]\omega_f = 0[/tex] is the final angular velocity (the tires come to a stop)

[tex]\alpha = -19.6 rad/s^2[/tex] is the angular acceleration

[tex]\theta[/tex] is the angular displacement

Solving for [tex]\theta[/tex],

[tex]\theta = \frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{0-(83.5)^2}{2(-19.6)}=177.9 rad[/tex]

And converting into revolutions,

[tex]\theta = \frac{177.9 rad}{2\pi}=28.3 rev[/tex]

c)

To find the time taken for the tires to stop, we use another suvat equation:

[tex]\theta = (\frac{\omega_i + \omega_f}{2})t[/tex]

where

[tex]\omega_i = 83.5 rad/s[/tex] is the initial angular velocity

[tex]\omega_f = 0[/tex] is the final angular velocity (the tires come to a stop)

[tex]\theta=177.9 rad[/tex] is the angular displacement

t is the time

And solving for t,

[tex]t=\frac{2\theta}{\omega_i + \omega_f}=\frac{2(177.9)}{83.5+0}=4.26 s[/tex]

d)

The distance travelled by the car during this time corresponds to the distance travelled by a point along the circumference of the tires.

We know that the angular displacement of one point of the tires during this time is

[tex]\theta=28.3 rev[/tex]

The radius of the tires is

r = 0.275 m

So the length of the circumference is

[tex]c=2\pi r=2\pi(0.275)=1.73 m[/tex]

Therefore, the total distance travelled by a point on the circumference (and therefore, by the car) is

[tex]d=\theta c = (28.3)(1.73m)=49.0 m[/tex]

e)

The initial velocity of the car corresponds to the initial linear velocity of a point along the edge of the tires, therefore it is given by

[tex]v_i = \omega_i r[/tex]

where

[tex]\omega_i = 83.5 rad/s[/tex] is the initial angular velocity

r = 0.275 m is the radius of the tires

Substituting the values, we find:

[tex]v_i = (83.5)(0.275)=23.0 m/s[/tex]

Learn more about angular motion here:

https://brainly.com/question/9575487

https://brainly.com/question/9329700

https://brainly.com/question/2506028

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