Question

At noon, ship A is 180 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 shipschanging at 4:00 PM?

Have to use differentiation to solve. Solving for I believe.

Answer 1

In 4hrs :
distance traveled by A = 4*40 = 140KM
distance traveled by B = 4*25 = 100KM
Angle between the line joining A and B with horizontal:
tanθ = 100/(180-140) = 2.5

θ=68.2 degrees

rate at which the distance between the ships changing at 4:00 PM is the magnitude of relative velocity between A and B:

velocity vectors:

V(A)= 40i

V(B)= 25j

relative velocity = Vr = V(A) - V(B) = 40i - 25j

magnitude of Vr =√(40^2 + 25^2) = 47.17m/s ....Ans

Answer 2

lets say the east-west direction is x and north south be y.

after 4 hours the distance between ships in x direction will be = 180 + 40*4 = 240 km

after 4 hours the distace between both in y direction is = 25 * 4 = 100 km

the distance between both the ship as a function of x and y direction is(S)

as given= 40 ,= 25

S =

dS/dt = = 2*240*40 + 2*100*25/2*260 =50.41 km/hr

dS/dt = 50.41 km/hr