Question

A ship leaves a port at noon and travels due west at 20 knots. At 6 PM, a second ship leaves the same port and travels northwest at 15 knots. How fast are the two ships moving apart when the second ship has traveled 90 nautical miles?

Answer 1

Answer:

$v = 12.44 Knots$

Explanation:

First ship starts at Noon with speed 20 Knots towards West

now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West

so after time "t" of 2nd ship motion the two ships positions are given as

$r_1 = 20(t + 6)\hat i$

$r_2 = 15(t)(cos45\hat i + sin45\hat j)$

now we can find the distance between two ships as

$x = \sqrt{(20(t + 6) - 10.6 t)^2 + (10.6t)^2}$

now we have

$x^2 = (120 + 9.4 t)^2 + (10.6 t)^2$

$x^2 = 200.72 t^2 + 14400 + 2256 t$

now we will differentiate it with respect to time

$2x\frac{dx}{dt} = 401.44 t + 2256$

here we know that

$t = \frac{90}{15} = 6 hours$

so we have

$x = 187.5$

now we have

$2(187.5) v = 401.44(6) + 2256$

$v = 12.44 Knots$