A car is travelling on an inclined road of inclination `30^(@)`, with a speed of 27 kmph. If the engine of the car is switched off how far up in the plane will car move before coming to rest, if the coefficient of friction between the tyre and the road is `0.4`.

Explanation

Angle of inclination `theta=30^(@)`
`u=27kmph=7.5m//s`
`mu_(k)=0.4`
Force opposing the motion of the car
`=mgsintheta+` Force of friction
`=mgsintheta+mu_(k)N`
`=mg sintheta+mu_(k)mg costheta`
`=mg(sintheta+mu_(k)costheta)`
Retardation of the car,
`a=g(sintheta+mu_(k)costheta)`
Final velocity `=v=0`
Distance travelled before coming to stop `=s=?`
`v^(2)=u^(2)+2as`
In this case a is negative
`0=7.5^(2)-2g(sintheta+mu_(k)costheta)s`
`s=(7.5^(2))/(2g(sintheta+mu_(k)costheta))=(7.5^(2))/(2xx9.8(sin30^(@)+0.4cos30^(@)))`
`=3.4m`