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at noon, ship a is 170 km west of ship b. ship a is sailing east at 40 km/h and ship b is sailing north at 25 km/h. how fast is the distance between the ships changing at 4:00 pm?



Answer :

If at noon , the ship A is 170 km west of ship B , then the distance between the ships changing at 4:00 p.m. is 28.85 kmph   .

The ship A is 170 km west of ship B and

ship A is sailing east at the speed of  40 km/h ;

ship B is sailing north at the speed of  25 km/h ;

let us consider "x" and "y" as initial position of ships A and B ;

let the distance between ship A and "x" = x ;

let the distance between ship B and "x" = y ;

let the distance between ship A and ship B = z ;

By using Pythagoras Theorem ;  z² = x² + y²  ;

differentiating w.r.t. "t" ,

we get ;

2z*dz/dt = 2x*dx/dt + 2y*dy/dt

on Dividing equation by 2 on both sides ;

z*dz/dt = x*dx/dt + y*dy/dt     .....equation(1)

given ; dx/dt = 40 km/hr  and   dy/dt = 25 km/hr

Now , When t = 4 hours,

since at noon ; the ship A is west of Ship B ;

we get ;   x = 40*4 = 170 - 160 = 10  ;  y = 25*4 = 100 and

z = √(x² + y²)  , putting the values of x and y ,

we get ;

z = √(x² + y²)  ;

z = √(10² + 100²)  ;

substituting the values ; we get

z = √100+10000

z = √10100

z = 100.49

putting the values of x , y and z , in equation(1) ,

we get ;

100.49*dz/dt = 10*40 + 100*25 ;

100.49*dz/dt = 400 + 2500

100.49*dz/dt = 2900

dz/dt = 2900/100.49

dz/dt = 28.85 kmph ;

Therefore , The distance between the ships is changing at 4:00 p.m. is 28.85 kmph .

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