This question is typical on some driver’s li- cense exams: A car moving at 40 km/h skids 17 m with locked brak?

9x longer or 9*17=153m reason is that breaking distance is proportional to square of velocity new velocity is three times higher (120/40=3) and 3 square is 9.

let's use the equation vf^2=v0^2 + 2ad to guide our answer in this equation, we know vf=final velocity =0 v0=initial velocity a=acceleration and d = distance traveled therefore, stopping distance is given by d = - v0^2/2a if we hold the acceleration constant (locking brakes in both cases) we see that stopping distance depends on the square of the initial velocity therefore, if you triple the initial velocity, the stopping distance increases by 3^2 or 9 times the new stopping distance is 9 x 17m = 153m

let's use the equation vf^2=v0^2 + 2ad to guide our answer in this equation, we know vf=final velocity =0 v0=initial velocity a=acceleration and d = distance traveled therefore, stopping distance is given by d = - v0^2/2a if we hold the acceleration constant (locking brakes in both cases) we see that stopping distance depends on the square of the initial velocity therefore, if you triple the initial velocity, the stopping distance increases by 3^2 or 9 times the new stopping distance is 9 x 17m = 153m