Question

Problem 1.78:

A ship leaves the island of Guam and sails a distance 300 km at an angle 40.0 ∘ north of west.

Part A: In which direction must it now head so that its resultant displacement will be 110 km directly east of Guam? (Express your answer as an angle measured south of east). theta = ________ degrees South of East.

Part B: How far must it sail so that its resultant displacement will be 110 km directly east of Guam?

______ km

Answer 1

first displacement :

d₁ = 300 km , 40 degree north of west

d_1x = X- component = - 300 Cos40 = - 229.81km               towards west

d_1y = Y- component = 300 Sin40 = 192.8 km               towards north

Net displacement ::

d_net = 110 km

d_netx = 110 km    towards east

d_nety = 0 km

Let the second displacement be d2 and d2x and d2y are its components

then Along X-direction

d_netx = d_1x + d_2x

110 = - 229.81 + d_2x

d_2x = 339.81 km

then Along Y-direction

d_nety= d_1y + d_2y

0 = 192.8 + d_2y

d_2y = - 192.8 km

d₂ = sqrt (d_2x² + d_2y²)

d₂ = sqrt (339.81² + (-192.8)²)

d₂ = 390.7 km

direction :

= tan^-1 (d_2y / d_2x) = tan^-1 (192.8 /339.81) = 29.6 degree south of east

B)

displacement = d₂ = 390.7 km