The intensity of sunlight hitting the surface of the earth on a cloudy day is about 0.50 kw/m2 assuming your pupil can close down to a diameter of 2.0 mm and that the average wavelength of visible light is 550 nm, how many photons per second of visible light enter your eye if you look up at the sky on a cloudy day

To determine the number of photons emitted per second, we need data on the total power of the system and the total energy of the photons. We determine the power by multiplying the total area to the intensity of the sunlight given. We do as follows:

Power = 0.50 kW / m^2 ( π (0.001^2)) = 1.57x10^-6 kW = 1.57x10^-3 W

The energy of the visible light can be calculated from the expression:

Energy = hc / wavelength

where h is the planck's constant and c is the speed of light

Energy = (6.63 x 10^-34) (3 x 10^8) / 550 x 10^-9
Energy = 3.62 x 10^-19 J

Number of photons per second =  1.57x10^-3 J/s / 3.62 x 10^-19 J
Number of photons per second = 4.34x10^15 

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. 

The answer is -6hope this