Question

A sailboat sails 2km due east, 5km $ 37^\circ $ south of east, and finally an unknown displacement. If the final displacement of the boat from the starting point is 6km due east, the third displacement is..........

Answer 1

Hint: The displacement of an object is the distance between the initial and final point of the sailboat. We will calculate the third displacement of the boat using direction vectors such that the net displacement will be 6 kilometers.

Complete step by step answer
Let us start by giving vector notation to the different displacements of the sail boat as given in the question. For the first displacement, the sail boat moves East which corresponds to the positive $ x $ direction which corresponds to the $ \hat i $ direction in the Cartesian system. So the displacement will be
 $ {S_1} = 2\hat i $
In the second displacement, the sail boat moves $ 37^\circ $ south of east which means the displacement will be
 $ {S_2} = 5(\cos 37^\circ )\hat i - 5(\sin 37^\circ )\hat j $ which can be written as
 $ {S_2} = 4\hat i - 3\hat j $
Let the displacement of the third vector be $ {S_3} $ . Since the net displacement of the boat is 6 kilometers due East, we can write the net displacement $ S $ as
 $ S = 6\hat i $
Since the individual displacements should add up to the net displacement, we can write
 $ S = {S_1} + {S_2} + {S_3} $
Substituting the value of $ {S_1},{S_2}\,{\text{and}}\,S $ as we derived above, we get
 $ 6\hat i = \hat i + 4\hat i - 3\hat j + {S_3} $
Solving for $ {S_3} $ , we get
 $ {S_3} = 3\hat j $ which is the third displacement.

Note
While calculating the displacement in the second displacement, the y-component will be negative since the boat is moving in the downwards direction i.e. South of East. Since we’ve been given the displacements of all the three different motions of the boat, they correspond to the distance between the final and the initial points of the boat and hence we can directly use the vector addition of displacements to calculate the net displacement.