Question

A ship leaves port at​ 12:30 pm and travels Upper N 55 degrees Upper E at the rate of 8 mph. Another ship leaves the same port at​ 1:30 pm and travels Upper N 30 degrees Upper W at the rate of 10 mph. How far apart in miles are the ships at​ 2:00 pm​ ?

Answer 1

Answer:

d=0.0167mille

Step-by-step explanation:

we must determine its position (X-Y).

ship 1

$v_{1}=8\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-12.5=1.5h\\d_{s1}=v_{1}*t_{t1}=8\frac{m}{h}*1.5h=12m$

ship 2

$v_{2}=10\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-13.5=0.5h\\d_{s2}=v_{1}*t_{t2}=10\frac{m}{h}*0.5h=5m$

component  x-y

$sin55=\frac{y_{1}}{h};y_{1}=12*sin55=9.82\\   cos55=\frac{x_{1} }{h};x_{1}=5*cos55=6,88$

$sin150=\frac{y_{2}}{h};y_{2}=5*sin150=2.5\\cos150=\frac{x_{2} }{h};x_{2}=5*cos150=-4.33$

$s_{1}=(6.88,9.82); s_{2}=(-4.33,2,5)$

we must find the distance S1-S2

$d_{s1-s2}=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}=\sqrt{(2.5-9.8)^{2}+(-4.33-6.88)^{2}}\\=\sqrt{(24.5)^{2}+(-11.21)^{2}}=26.94m$

but the units are requested to be miles

d=26.94m*mille/1,609.34m=0.0167mille