a = 30/12 = 2.5 m/s^2
F-200 = m a = 1100(2.5) =2750 N
F = 2950 N
3.16. An 1100-kg car accelerates from 0 to 30 m/sec in 12 sec. If a frictional force of 200 N opposes the motion, what forces must the wheels exert backward on the pavement to cause this acceleration?
1 answer