CONTROL CIRCUIT : Question Regarding Voltage and Resistance
we've got a capacitor and a series resistor and then point c

we've got a graph of "voltage at point c" that looks like it might be  logarithmic.

from the text it can be understood that when voltage at point c drops low enough the deflector is activated. the deflector then operates autonomously for 2 seconds deploying and the retracts (presumably taking approximately another 2 seconds but that's not important)

we're told point C is connected to a voltage activated trigger 

to have the deflector fully deployed at 3 seconds we need to delay for only 1 second as there are two seconds of delay inherent in the deflector.
that means deploy at 6V 

to increase the delay by mofiying the external components the easiest answer is to increase the capacitance of the capacitor, this will make the discharge take longer (I'm assuming the capacitor sis charged to a fixed voltage - this seems like a reasonable assumption)

the effect of the resistor on the voltage at point C is more complicated.

assuming the bottom end of the capacitor is held at a constant voltage during the discharge. We aren't told this, but from the graph it seems likely, (and we aren't given any other point to measure voltage from!)
point C behaves somwwhat like a resistor to ground: the more current is flowing into it the higher the voltage.

looking carefully at the graph reveals that it is not exponetial decay.  in the first second the voltage halves, but in the next second it doesn't.
so point C isn't pure resistance to ground... looking even more closely, I am unable to fit any exponential curve to that graph, thus the terminal behind point C can't be treated as a fixed thevenin equivalent circuit. 

The graph looks approimately hyperbolic, I can't immagine what circuitry or electrical behaviour behind terminal C would be required to get that effect.

Either this is a trick question, or it is very badly written.