Answer: After 4.55 hours the ships will be nearest to each other, and this distance would be 54.54 km.
Step-by-step explanation:
Since we have given that
Distance between ship A and ship B = 100 km
Speed at which Ship A is sailing west = 12 km/hr
Speed at which Ship B is sailing south = 10 km/hr
Since they are sailing opposite to each other, so their relative speed would be
[tex]12+10=22\ km/hr[/tex]
So, Time taken is given by
[tex]time=\dfrac{Distance}{Speed}=\dfrac{100}{22}=4.55\ hrs[/tex]
Distance would be
[tex]\dfrac{x}{12}=\dfrac{100-x}{10}\\\\10x=12(100-x)\\\\10x=1200-12x\\\\10x+12x=1200\\\\ 22x=1200\\\\x=\dfrac{1200}{22}\\\\x=54.54\ km[/tex]
Hence, After 4.55 hours the ships will be nearest to each other, and this distance would be 54.54 km.
Answer:
at 4:55 pm
About 64 km
Step-by-step explanation:
Let t be time in hours taken by ship A to point B and ship B takes time t hours to point A
Speed of ship A=12 km/h
Speed of ship B=10 km/h
Distance=[tex]speed\times times[/tex]
Distance travel by ship A=12 t
Distance travel by ship B =10 t
OA=10 t
OC=100 km
OB=OC-BC=100-12 t
Using pythagorous theorem
[tex](Hypotenuse)^2=(perpendicular)^2+(base)^2[/tex]
[tex]AB^2=(10t)^2+(100-12 t)^2[/tex]
[tex]D=(10t)^2+(100-12 t)^2[/tex]
The distance between two ships is minimum
Differentiate w.r.t time
[tex]\frac{dD}{dt}=200t+2(100-12t)\cdot (-12)[/tex]
[tex]\frac{dD}{dt}=0[/tex]
[tex]200 t-24(100-12t)=0[/tex]
[tex]488t-2400=0[/tex]
[tex]t=\frac{2400}{488}=4.92 hours[/tex]
Again differentiate w.r.t
[tex]\frac{d^2D}{dt^2}=200-24(-12)=200+288=488 > 0[/tex]
Hence, the distance is minimum
The time will be 4:55 pm
Substitute t=4.92
Then AB=[tex]\sqrt{(49.2)^2+(40.96)^2}[/tex]
AB=64.01 Km
AB=64 km
Hence, the distance between two ships =64 km (about)
