a 1240 kg automobile stops in a distance of 41m. if initial velocity was 22.5m/s, what force did the road exert on the car to stop it? Answer pls... Thank You
V^2 = Vo^2 + 2a*d.
0 = 22.5^2 + 2a*41, a = ?.
"a" will be negative which means that the acceleration opposes the motion.
F = M*a = 1240*a.