vf^2=vi^2+2 a d
but a=force/mass=-mu*mg/m=-mu*g
vf=0 change 105km/hr to m/s 29.16 ? check that.
0=29.16^2=mu*g*d
solve for d
A 1500 kg automobile travels at a speed of 105 km/h along a straight concrete highway. Faced with an emergency situation, the driver jams on the brakes, and the car skids to a stop.
(a) What will be the car's stopping distance for dry pavement (µ = 0.85)?
(b) What will be the car's stopping distance for wet pavement (µ = 0.60)?
1 answer
1 answer