Asked by Casey
At 7:00 am, one ship was 60 miles due east from a second ship.If the first ship sailed west at 20 mph, and the second ship sailed southeast at 30 mph, at what time are they closest together?
Answers
Answered by
Steve
Assume the ships started at A=(60,0) B=(0,0)
at time x hours,
A is at 60-20x
B is at (21.2x,-21.2x)
the distance d is thus given by
d^2 = (60-20x-21.2x)^2 + (-21.2x)^2
2d dd/dx = 2(60-41.2x)(-41.2) + 2(-21.2x)(-21.3)
we can forget about the denominator and just set the numerator = 0, to get dd/dx = 0 when x = 1.151, = 1:09
So, the ships are closest at 8:09 am
at time x hours,
A is at 60-20x
B is at (21.2x,-21.2x)
the distance d is thus given by
d^2 = (60-20x-21.2x)^2 + (-21.2x)^2
2d dd/dx = 2(60-41.2x)(-41.2) + 2(-21.2x)(-21.3)
we can forget about the denominator and just set the numerator = 0, to get dd/dx = 0 when x = 1.151, = 1:09
So, the ships are closest at 8:09 am
Answered by
Ekwé
how did you derive where B is at where did the answer 21.2 come from?
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