The driver of a car traveling at 42. 2 m/s applies the brakes and undergoes a constant deceleration of 0. 99 m/s 2. 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. 14 m?.

Respuesta :

Before the car stop, each tire makes 149.14 revolutions. The rate of change in the velocity of the object is called acceleration.

What is acceleration?

The rate of change in the velocity of the object is called acceleration. The distance can be calculated by the formula,

[tex]\Delta x = \dfrac {v_f - v_i}{2d}[/tex]

Where,

[tex]\Delta x[/tex] - distance travelled

[tex]v_f[/tex] - final velocity

[tex]v_i[/tex] - initial velocity = 42.2 m/s

[tex]d[/tex] - decceleleration = [tex]0. 99 \rm \ m/s^2[/tex]

Put the values in the formula,

[tex]\Delta x = 20.88 \rm \ m[/tex]

Now, from the relation between angular and linear displacement,

[tex]\Delta \theta = \dfrac {\Delta x }r[/tex]

Where,

[tex]\Delta \theta[/tex] - angular displacement

r - radius = 0. 14 m

Put the values in the formula,

[tex]\Delta \theta = \dfrac { 20.88 }{0.14}\\\\\Delta \theta = 149.14 \rm \ revolutions[/tex]

Therefore,  before the car stop, each tire makes 149.14 revolutions.

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