the boy's acceleration toward the girl is 2.08 m/s^2.
explanation:
according the newton's third law, the force causing the girl accelerate towards the boy will be me with equal counter-force causing the boy to accelerate towards the girl. these forces will be of equal magnitude and allows us to determine the boy's acceleration:
![f_{boy} = f_{girl}\\f_{girl} = m_{girl}\cdot a_{girl} = 37kg\cdot 3.2 \frac{m}{s^2}=118.40n\\\implies f_{boy} = 118.40n\\a_{boy} = \frac{f_{boy}}{m_{boy}}=\frac{118.4n}{57kg}=2.08\frac{m}{s^2}](/tex.php?f=f_{boy} = f_{girl}\\f_{girl} = m_{girl}\cdot a_{girl} = 37kg\cdot 3.2 \frac{m}{s^2}=118.40n\\\implies f_{boy} = 118.40n\\a_{boy} = \frac{f_{boy}}{m_{boy}}=\frac{118.4n}{57kg}=2.08\frac{m}{s^2})
the boy's acceleration toward the girl is 2.08 m/s^2.