When a car moving with a speed of `36km//h` reaches an upwards inclined road of angle `30^(@)` its engine is switched off If coefficient of frition involved is `0.1` how much distance will the car move before coming to rest `Given `g = 10m//s^(2)`.

Answer

`8.52m`

Explanation

Here, `u = 36km//h =10m//s, 0 =30, mu = 0.1`
`s =? upsilon = 0`
While moving up the incline, retardation of car
`a = g (sin theta + mu cos theta)`
`=10 (sin 30 + 0.1 cos 30) = 5.866m//s^(2)`
From `upsilon^(2) - u^(2) = 2 as`
`s = (upsilon^(2) - u^(2))/(2a) = (0 -10^(2))/(2 (-5.866)) = 8.52 m`.