Two ships A & B are moving at constant velocities. At t=0 the position of ship A

Question

Two ships A & B are moving at constant velocities. At t=0 the position of ship A is 2.41 miles east and 3.14 miles north of ship B. Ship A has a velocity of 36.2 mph and is headed south. Ship B has a velocity of 17.9 mph and is at an angle of 40 degrees north of east. (a)Determine the relative velocity, V_B/A, of B with respect to A. (b)Determine r_B/A(t), the relative position of B with respect to A as a function of time. (c)Determine the distance between A and B after 3 minutes. (d) Determine the minimum distance between the ships and the time when this occurs.

Explanation / Answer

you have found out v_B/A = -13.7 i + 47.7 j

b) Now r_B/A = V_B/A x t

= (-13.7 i + 47.7 j) x t (as dr/dt =- V so, r = v.dt)

c) distance at t = 3 minutes = 3/60 hr =

( -13.7 i + 47.7 j ) x 3/60 = - 0.685 i + 2.385 j miles {nOTE : to get distance in feet multiply the distance with 5280 as 1 milke = 5280 foot}

d) let the coordinate of A be arbitarily assigned (0 i + 0j).

so with respect to A, B is at a distance of (-13.7 i + 47.7 j) x t

so, dstance between them = S =sqrt { (-13.7 t - 0)^2 + (47.7t - 0)^2}

so, S^2 = (-13.7.t)^2 + (47.7 .t)^2

for minimum distance DS/dt = 0

diiferentitain above euation with time :

S.dS/dt = 2 x(-13.7)^2.t + 2 x (47.7)^2t = 0

so, t = 0 minimum distance occurs.

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