no no don't be sorry, my explanation was maybe too vague.
here is a diagram;
time = 0 is the instant the scenario starts, we don't need to worry about anything before that.
let theresa have constant speed 'v' ms-1, this means no matter what the time is theresa will be running at 'v'
now, since magda is running with an acceleration of 1 ms-2 this means that at;
t = 0, v = 0
t = 1, v = 1
t = 2, v = 2
and most importantly, at t=v, v = v
Now this point is important because once magda starts running at speeds more than v, it will be physically impossible for theresa to catch up since madga will be running faster.
Since the question tells you that theresa
just catches magda this means she reaches madga at time = v, ie the last possible second before magda starts running at a faster speed than theresa.
got everything so far?
so now we find the area under both curves/lines from time = 0 to time = v.
However, on top of this, we must remember that theresa has to run an extra 4.5m in this time period than magda to catch her.
So "distance theresa runs + 4.5 = distance magda runs" between t=0 and t=v
re wording this slightly we get to;
\DISPLAYSTYLE \text{Area under theresa's curve + 4.5 = Area under magda's curve , between t=0 and t=v}