so we are looking for the force of braking (ie, the max deacceleration allowed).
vf=vi+at
vf-vi=at
Force=mass*acceleration, so
acceleration=frictionforce/mass=mass*9.8N/kg*coefficentfriction/mass=9.8*coefficent
so if deacceleration is constant, regardless of speed.
and you Double (70 to 140) the change in speed, then time must be doubled (to 16seconds)
An 3000-kh car traveling at 70 m/s takes 8 m to stop under full braking. The same car under similar road conditions, traveling at 140 m/s, takes
2 answers
the braking force dissipates the kinetic energy of the vehicle
if the velocity of the vehicle is doubled , the KE is quadrupled ... KE = 1/2 m v^2
... so the brakes must dissipate four times the energy
if the braking force is constant , it takes four times as long to stop
if the velocity of the vehicle is doubled , the KE is quadrupled ... KE = 1/2 m v^2
... so the brakes must dissipate four times the energy
if the braking force is constant , it takes four times as long to stop
1 answer