Question

At 3 P.M, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 7 P.M.? (Round your answer to one decimal place.)

Answer 1

 Starting position: 

Let B start at origin(O) & A is at (-150 ,0) ie 150 km west of the origin . At 3pm , t= 0 
A is changing its position at ( 150-35t) km/h 
& B is changing its position at 25t km/h 

Therefore 

AB^2 = OA^2 + OB^2 /// Pythagoras Theorem 
= (150-35t)^2 + ( 25t)^2 
= 22 500 - 10 500 t + 1225t^2 +625t^2 
= 22 500 - 10 500t + 1850t^2 
AB = ( 22 500 - 10 500t + 1850t^2)^1/2 

d (AB) /dt = 1/2 * ( 22 500 - 10 500t + 1850t^2)^-1/2 * ( - 10 500 + 3700t) 
= ( - 5250 + 1850t) / ( 22 500 - 10 500t + 1850t^2)^1/2 

At 3pm , t=0 
At 7pm , t= 4 

So d (AB) /dt = ( -5250 + 1850*4) / ( 22500 - 10500*4 + 1850*4^2)^1/2 
= 2150 / 100.5 
= 21.4 km/hr

The distance between the ships changing at 7 P.M. is with a speed of 21.4 km/hr.

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