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.)

 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.

I hope my answer has come to your help. Thank you for posting your question here in .

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

What is the last question i don’t see it?