Answer
(a) The man's acceleration's magnitude is $0.55m/s^2$. Its direction is eastward.
(b) The woman's acceleration's magnitude is $0.94m/s^2$. Its direction is westward.
Work Step by Step
We take east to be the positive direction, so the force exerted on the man is $+45N$.
(a) $F=+45N$ and $m_{man}=82kg$. The acceleration of the man is $$a=\frac{F}{m_{man}}=+0.55m/s^2$$
The man's acceleration's magnitude is $0.55m/s^2$. Its direction is eastward.
(b) According to Newton's 3rd Law, as the woman exerts the force on the man, there is an oppositely directed force of equal magnitude back on the woman. In other words, there is a force $F_{reaction}=-45N$ on the woman.
$m_{woman}=48kg$. The acceleration of the woman is
$$a=\frac{F_{reaction}}{m_{woman}}=-0.94m/s^2$$
The woman's acceleration's magnitude is $0.94m/s^2$. Its direction is westward.
