A1300 -kg car is pushing an out-of-gear 2160 -kg truck that has a dead battery. when the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4420 n . the rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 770 n , including the "friction" of turning the truck's drivetrain. what is the magnitude of the force the car applies to the truck?

Ft = 3650 N

Explanation:

The car exerts a force on the truck and truck has a friction force, using Newton's second law we can find the net force is the difference between the two

    Ft = F-fr

    Ft = 4420-770

    Ft = 3650 N

Directed from the car to a truck .  The system acceleration is

   Ft = (m₁ + m₂) a

   a = Ft / (m₁ + m₂)

   a = 3650 / (1300 +2160)

   a = 1.05 m / s²

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)