The driver of a car cannot withstand an acceleration of more that 3g. g is the acceleration due to gravity and equal to 9.8 m/s^2; 3g is an acceleration equal to 29.4 m/^2.
If the driver is initially traveling at 120 km/h and needs to stop at a point. The deceleration when the brakes are applied cannot exceed 29.4 m/s^2. Let the distance where the brakes are applied from the point where he intends to stop be S m.
Using the formula `v^2 - u^2 = 2*a*s`
`0^2 - (120*(1000/3600))^2 = -2*29.4*S`
=> S = 18.89 m
The brakes should be applied at least 18.89 m from the point where the driver wishes to stop.