Answer:
After 21.6 minutes after 5:00 pm boats are closest to each other.
Step-by-step explanation:
Speed of the first boat = 20 [tex]\frac{km}{hr}[/tex]
Let after time t the distance between the boats is minimum.
Distance traveled by the first boat in time t is = 20 t ------- (1)
Speed of the second boat = 15 [tex]\frac{km}{hr}[/tex]
Distance traveled by the second boat in time t is = 15 t ------- (2)
From the diagram
Distance AB = 15 - 15 t --------- (3)
Distance between the boats is at time t is AC. And it is calculated by
[tex]AC^{2} = AB^{2} + BC^{2}[/tex]
[tex]AC^{2} = (15 - 15 t) + ( 20 t)^{2}[/tex]
Differentiate above equation with respect to t we get,
[tex]AC'^{2} =[/tex] 2 (15 - 15 t) (-15) + 2 (20 t) (20)
[tex]AC'^{2} =[/tex] - 450 + 1250 t
for the distance to be minimum [tex]AC'^{2} =[/tex] 0
⇒ - 450 + 1250 t = 0
⇒ t = [tex]\frac{450}{1250}[/tex]
⇒ t = 0.36 hours
⇒ t = 21.6 minutes
Therefore the distance between the boats after 21.6 minutes after 5:00 pm is minimum.
