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

d=0.0167mille

Explanation:

we must determine its position (X-Y).

ship 1

`v_(1)=8(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(m)/(h)*1.5h=12m`

ship 2

`v_(2)=10(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(m)/(h)*0.5h=5m`

component x-y

`sin55=(y_(1))/(h);y_(1)=12*sin55=9.82\\   cos55=(x_(1) )/(h);x_(1)=5*cos55=6,88`

`sin150=(y_(2))/(h);y_(2)=5*sin150=2.5\\cos150=(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