The capacitor in an rc circuit is discharged with a time constant of 8.00 ms . part a at what time after the discharge begins is the charge on the capacitor reduced to half its initial value? express your answer with the appropriate units

You need to be more specific.  i really don't know what a "motion map" is.you can draw a graph of an object's position, distance, displacement, speed,velocity, or acceleration versus time, or you can draw a graph of the object's x-position versus its y-position and not give any information about time at all.if the graph presents position or displacement against time,  then constant velocity is a straight line . . i.e. a line that has  constant slope.if the graph shows speed against time, then constant velocityis a horizontal piece of line.if the graph shows acceleration against time, then constant  velocity is a piece of line that lies on the x-axis . . it's value is zero. 

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.