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,
Along X-direction, the force is
Friction acts in the opposite way along the x-axis
⇒-μ
⇒-μmg =
⇒ 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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