Consider the 65.0 kg ice skater being pushed by two others shown in the figure?

c isn't 0.468 either

Mar 12, 2017

$A = \frac{F}{m} = \frac{30.44}{65.0} = 0.468$ $m {s}^{- 2}$

That is the correct answer in physics. So much the worse for whoever wrote the question is not the answer they have set!

Explanation:

The initial velocity is irrelevant, since velocity is relative anyway. The net force acting is $30.44$ $N$ on a mass of $65$ $k g$, so the acceleration has to be $0.468$ $m {s}^{- 2}$.

The number of significant digits is even correct.

Edit: The reminder about friction is interesting. We are told she is wearing steel-bladed skates on ice. Was there earlier information on frictional coefficients for that situation? If such, there would be a small force opposing the accelerating force, so her acceleration would be less. You could calculate that as follows:

F_{frict} = \muF_{norm} = \mumg

You would subtract that frictional force from the applied force before moving to the $a = \frac{F}{m}$ step.