SOLUTION: 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 3

Question 187755This question is from textbook
: 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.
This question is from textbook

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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.
:
From the given information we can say:
t1 = t2 + 12 (seconds)
:
When destruction occurs, both rockets will have traveled the same distance
write a distance equation from this fact: Dist = speed * time
:
2000(t1) = 3200(t2)
t1 =
t1 = 1.6(t2)
Substitute (t2-12) for t1; find t2
t2 + 12 = 1.6t2
12 = 1.6t2 - t2
12 = .6t2
t2 =
t2 = 20 seconds
then
t1 = 20 + 12 = 32 seconds
;
:
Check solution by ensuring they both traveled the same distance
2000 * 32 = 64000 ft
3200 * 20 = 64000 ft also

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