Question 2804
Suppose in time t the second car will be 30 miles ahead of first
Let the distance covered by the slower car be x miles in t hrs
so the distance covered by the fast car will be x+30 in t hrs

so time taken by the first car will be = distance/speed = x/45
and time taken by the second car will be  = (x=30)/53

But the time taken is same t hrs
so the equation will be
x/45 = (x+30)/53
=> 53x = 45x + 45*30
=> 8x = 45*30
=> x = 45*30/8
=> x = 168.75

So the time will be 168.75/45 = 3.75
So in 3.75 hrs the second car will be 30 miles ahead of first