Question

A 30 kg girl and a 40 kg boy stand on roller skates, face one another, and

push off each other. If the boy accelerates backward with an acceleration
of 6 m/s2, what will be the acceleration of the girl?

Answer 1

By using Newton's Third Law and Newton's Second Law, we determined that, when a 30 kg girl and a 40 kg boy push off each other with the boy accelerating at 6 m/s², the acceleration of the girl will be 8 m/s² in the opposite direction.

Step-by-step explanation:

The question pertains to Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. When the girl and the boy push off each other, the force the boy applies to the girl is equal in magnitude but opposite in direction to the force the girl applies to the boy. Thus, by Newton's Second Law (F = ma), we can find the acceleration of the girl by knowing the acceleration of the boy and the masses of both individuals.

For the boy:

• Force applied by boy (F_boy) = Mass of boy (m_boy) × Acceleration of boy (a_boy)
• F_boy = 40 kg × 6 m/s² = 240 N

This force is also applied to the girl, but in the opposite direction:

• Force applied to girl (F_girl) = 240 N

Using the girl's mass:

• Acceleration of girl (a_girl) = Force applied to girl (F_girl) / Mass of girl (m_girl)
• a_girl = 240 N / 30 kg
• 
  `a_girl` ²

Therefore, the acceleration of the girl is 8 m/s² in the direction opposite to that of the boy.