One athlete in a race running on a long, straight
track with a constant speed v, is a distance d behind a sec-
ond athlete running with a constant speed u. (a) under
what circumstances is the first athlete able to overtake the
second athlete? (b) find the time t it takes the first ath-
lete to overtake the second athlete, in terms of d, v,, and
vy. (c) at what minimum distance d, from the leading
athlete must the finish line be located so that the trailing
athlete can at least tie for first place? express d, in terms
of d, v1, and v, by using the result of part (b).

Dis the

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. 

Make them both to where they have the denominator of 18that makes 3/2 equal 27/18 and 1/9 equal 2/18so now you have  27/18x+5=2/18subtract 5 from the other sideso now you have 27/18x+2/18-5then subtract 2/18 from 27/18 to get 25/18x=-5then divide 25/18 by -5 and your answer will be  -5/18