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A motorboat is observed to travel10kmph relative to earth in the direction 37deg north of east.If the velocity of boat due to the wind only is 2kmph westward and that due to current only is 4kmph southward,what is the magnitude and direction of the velocity of the boat due to its own power.
East - i vector North - j vector.
Vel of boat = vel of itself + vel of wind + vel of current.
Vb = Vit + Vw + Vr
10(cos 37 i + sin 37 j) = Vit + (-2 i) + (-4 j)
8 i + 6 j = Vit + (-2 i) + (-4 j)
Vit = 10 i + 10 j
Boat travels at 14.14 kmph 45deg north of east due to its owm power.
vel. of boat w.r.t wind Vbw = -2j
vel. of boat w.r.t water current Vbc = -4i
given vel. of boat w.r.t earth(water current + wind): Vb(w+c) = 6i + 8j
so, Vb - Vw = -2j .............(1)
Vb - Vc = -4i .............(2)
Vb - V(w+c) = 6i + 8j . ............(3)
Let vel. of boat alone due to its power is = (ai+bj)
From above (1) + (2) -(3) gives
Vb = -10.(i + j)
The magintude of vel. of boat = 14.14 kmph
direction is 45 degrees south west
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