Boat A leaves a dock headed due east at 1pm travelling at a speed of 10 mi/hr. At 6pm ,boat B leaves the same dock travelling due south at a speed of 25mi/hr.

Question

Boat A leaves a dock headed due east at 1pm travelling at a speed of 10 mi/hr. At 6pm ,boat B leaves the same dock travelling due south at a speed of 25mi/hr.
 Find an equation that represents the distance d in miles between the boats and any time t in hours for t ≥ 5 , using that t =0 corresponds to the time that boat A leaves the dock.
thank you

oobleck · Answer

distance = speed * time, so at time t, you have A: 10t B: 25(t-5) so the distance z is z^2 = (10t)^2 + (25(t-5)^2 = 725t^2 - 6250t + 15625 z = 5√(29t^2 - 250t + 625)

Anonymous · Answer

east is +x direction south is -y direction x of boat A = 10 (t - 1 ), y of boat 1 remains 0 x of boat B = 0, y = -25 (t - 6) d = sqrt (x^2 + y^2) d^2 = 100 (t^2-2t+1) + 625 (t^2 -12 t + 36) continue