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The driver of a car traveling at 36.4 m/s applies the brakes and undergoes a constant
deceleration of 0.94 m/s2

How many revolutions does each tire make
before the car comes to a stop, assuming that
the car does not skid and that the tires have
radii of 0.35 m?
Answer in units of rev

Respuesta :

Each tire make  before the car comes to a stop is 322 revolutions.

Explanation:

Given that,

Initial speed of the car, u = 36.4 m/s

Finally, it stops, v = 0

Deceleration of the car, [tex]a=-0.94\ m/s^2[/tex]

Radius of the tire, r = 0.35 m

Let d is the displacement of the car. It can be calculated using third equation of kinematics as :

[tex]d=\dfrac{v^2-u^2}{2a}[/tex]

[tex]d=\dfrac{-(36.4)^2}{2\times -0.94}[/tex]

d = 704.76 meters

The distance covered in 1 revolution,

[tex]x=2\pi r[/tex]

[tex]x=2\pi \times 0.35[/tex]

x = 2.19 m

Let n are the number of revolution, so,

[tex]n=\dfrac{704.76}{2.19}[/tex]

n = 321.8

or

n = 322 revolutions

So, each tire make  before the car comes to a stop is 322 revolutions. Hence, this is the required solution.

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