@TommyTman
Question stem indicates a component of it's
acceleration is in the x-direction following the application of a 3rd force. The sum of the forces along the x-axis is negative in this case. Recall that Ftot = Fa + ... + Fn = ma.
While the RBC may still be traveling in the +x direction, it is accelerating in the opposite direction (i.e. decelerating in the +x direction).
In response to
@NITRAS , The question stem does not indicate anything as to the nature of the y-component of acceleration after the 3rd force has been applied. The y-direction force components do not necessarily cancel out, but we do know that the net force could be in quadrants 2,3, or along the x-axis because each would have a component of force in the -x direction.