Question

If the coefficient of friction between road and tires on a rainy day is 0.109, what is the minimum distance in which the car will stop

Answer 1

The car will halt at a minimum distance of 98.57 meters on a rainy day having a friction coefficient of 0.109.

Calculation of the minimum distance-

Provided that :

• the speed of the car = 52 km/h

                                  = 52 x 0.278 m/s = 14.45 m/s

• Friction coefficient, μ = 0.109

the regular force exerted on the car,

$F_{N} = mg$

Along X-direction, the force is

$F_{x} = ma_{x}$

Friction acts in the opposite way along the x-axis

$-f_{friction} = ma_{x}$

⇒-μ$F_{N} = ma_{x}$

⇒-μmg = $ma_{x}$

⇒$a_{x} =$ μg

Utilizing the motion equation-

v² = u²+ 2 .a. s

The final speed, v=0 m/s

⇒0² = (14.45)² - 2 μg .s

⇒2 * 0.109 *9.8 *s = (14.45)² = 208.8

⇒s = 208.8 / 2.14 =  97.57 m

It is concluded that the car will halt at a minimum distance of 98.57 meters.

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