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 ?
Accepted Solution
A:
Answer:d=0.0167milleStep-by-step explanation:we must determine its position (X-Y).ship 1[tex]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[/tex]ship 2[tex]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[/tex]component x-y[tex]sin55=\frac{y_{1}}{h};y_{1}=12*sin55=9.82\\ cos55=\frac{x_{1} }{h};x_{1}=5*cos55=6,88[/tex][tex]sin150=\frac{y_{2}}{h};y_{2}=5*sin150=2.5\\cos150=\frac{x_{2} }{h};x_{2}=5*cos150=-4.33[/tex][tex]s_{1}=(6.88,9.82); s_{2}=(-4.33,2,5)[/tex]we must find the distance S1-S2[tex]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[/tex]but the units are requested to be milesd=26.94m*mille/1,609.34m=0.0167mille