A car is going at

The maximum force of friction is developed between the surface of road and the car’s tyres when hard breaks are applied.

So, maximum frictional force = μR

From the free body diagram,

R − mg cos θ = 0
⇒ R = mg cos θ (i)
and
μR + ma − mg sin θ = 0 (ii)
⇒ μ mg cos θ + ma − mg sin θ = 0

⇒ μg cos θ + a -10(12) =0

⇒ a=5- {1-(2)} ×10(/2)

⇒ a = -2.5 m/s²

When brakes are applied, car will deaccelerate by 2.5 m/s²

Distance s= 12.8 m
initial velocity u = 6 m/s
∴ Velocity at the end of incline

ν=

v=10 m/s=36 km/h

Hence, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h