In a test of a heat-seeking rocket, the target rocket is launched at 2000 ft/sec and the heat-seeking rocket is launched along the same flight path 12 seconds later at a speed of 3200 ft/sec. Find the times t1 and t2 of flight of the rockets until the heat-seeking rocket destroys the first rocket.

Use the equation

X = V_0T - \frac 12 GT^2

Where X is the distance traveled, Vo is the initial speed, T is the time and G is the acceleration due to gravity. The distance traveled must be the same, so you can equate the distances traveled.

You also know that T2 = T1 + 12 seconds

I think it is just meant to be an exercise in d=rt.

2000t=3200(t-12)

Solve for t.

I came up with:

2000t=3200t-3200*12
-1200t=-38400
t=32 sec

Thank you

Yes. That means the faster rocket will have been in the air for 20 seconds and the one gwetting destroyed for 32 seconds.