A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle `30^0` figure. `1/(2sqrt3)`. Shwo tht no matter how hard the driver applies the brakes, tehcar will reach the bottim with a speed greater than 36 km/hr. Take `g=10 m/s^2`

Explanation

Hardest brake indicates that maximum force of friction is developed between car's tyre and road.
Maximum frictioN/Al force `=muR`
From the free body diagram,
`R-mgcostheta=0`
`rarr R=mg costheta`……..i
`and muR+ma-mgsintheta=0` ………ii
`rarr mu mg cos theta+ma-mg sin theta=0`

`rarr mg cos theta +a-10xx(1/2)=0`
`rarr a=5-[1-2sqrt3]xx10(sqrt3/2)`
`=5-2.5=2.5m/s^2`
`S=12.8 m/s^2 `
`u=6 m/s, velocity at the end of incline
`v=sqrt(u^2+2as)`
`=sqsrt(6^2+2(2.5)(12.8))`
`=sqrt(36+64)
`=10 m/s=36 km/hr
Hence, the harder the driver applies the brakes,m the lower will be the velocity of the cr when it reaches the ground i.e. at `36 km/hr`.