At a time of t after t=0 , the southward distance covered by A is 12t miles, and the distance covered by B is 8t
Let the distance between them be d
I see a right - angled triangle.
a) d^2 = (8t)^2 + (12-12t)^2
b)
dd/dt = 2(8t)(8) + 2(12-12t)(-12)
= 0 for a min d
128t -288 + 288t = 0
416t = 288
t = 288/416 = 9/13 = .6923
they will be closest at 9/13 hours
= .6923 hrs
= 41.5 minutes
check:
at t = .6923 , d = 6.65640
take a value of t slightly higher and lower than .3923
t = .69 , d = 6.65648 , a bit farther than at t = .6923
t = .70 , d = 6.657 , a bit farther than at t = .6923
My answer is correct
At t=0, ship A is 12 miles due north of ship B. Ship A travels 12 miles/hour due south, while ship B travels 8 miles/hour due east.
a. Write a function for the distance between the two ships.
b. At what time are the two ships closest?
2 answers
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I appreciate it!