Question

Three ships A, B & C are in motion. Ship A moves relative to B with speed $v$ towards North East. Ship B moves relative to C with speed $v$ towards the North-West. Then relative to A, C will be moving towards-
(A) North
(B) South
(C) East
(D) West

Answer 1

Hint To answer this question, we need to apply the formula of the relative speed of an object with respect to another object. Writing the relative speeds in the form of the unit vectors and manipulating the equations formed, we can determine the direction of the required relative speed.

Formula Used: The formula which is used in solving this question is given by
$\Rightarrow {v_{21}} = {v_2} - {v_1}$ , here ${v_{21}}$ is the relative speed of object $1$ with respect to the object $2$ , ${v_1}$ and ${v_2}$ are the respective speeds of the object $1$ and the object $2$ with respect to the ground.

Complete step by step answer
Let us take the unit vector \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} in the East direction, and the unit vector \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} in the North direction.
In terms of the unit vectors assumed, the unit vector in the North East direction can be taken as
\Rightarrow {\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{n} _1} = \dfrac{{\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }} …………………..(1)
Also, the unit vector in the North West direction can be taken as
\Rightarrow {\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{n} _2} = \dfrac{{ - \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }} …………………..(2)
Let ${v_A}$ and ${v_B}$ be the respective speeds of the ship A and the ship B with respect to the ground.
We know that the relative speed of a body with respect to another body is given by the formula,
$\Rightarrow {v_{21}} = {v_2} - {v_1}$
So, the speed of ship A relative to the ship B is given by
$\Rightarrow {v_{AB}} = {v_A} - {v_B}$ …………………..(3)
Also, the speed of ship B relative to the ship C is given by
$\Rightarrow {v_{BC}} = {v_B} - {v_C}$ …………………..(4)
According to the question we have
\Rightarrow {v_{AB}} = v{\text{ }}{\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{n} _1} and
\Rightarrow {v_{BC}} = v{\text{ }}{\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{n} _2}
From (1) and (2) we get
\Rightarrow {v_{AB}} = v{\text{ }}\left( {\dfrac{{\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right)
{v_{BC}} = v\left( {\dfrac{{ - \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right)
So from (3) and (4) we get by substituting,
\Rightarrow {v_A} - {v_B} = v{\text{ }}\left( {\dfrac{{\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right) …………………..(4)
\Rightarrow {v_B} - {v_C} = v\left( {\dfrac{{ - \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right) …………………..(5)
Now, the relative speed of the ship C with respect to the ship A will be given by
$\Rightarrow {v_{CA}} = {v_C} - {v_A}$ …………………..(6)
Adding eq (5) with eq (4) we get
\Rightarrow {v_A} - {v_B} + {v_B} - {v_C} = v{\text{ }}\left( {\dfrac{{\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right) + v\left( {\dfrac{{ - \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right)
On taking the velocity common,
\Rightarrow {v_A} - {v_B} + {v_B} - {v_C} = v{\text{ }}\left[ {\left( {\dfrac{{\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right) + \left( {\dfrac{{ - \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right)} \right]
On simplifying the LHS and the RHS, we get
\Rightarrow {v_A} - {v_C} = v{\text{ }}\left( {\dfrac{{2\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} }}{{\sqrt 2 }}} \right)
Hence we have,
\Rightarrow {v_A} - {v_C} = \sqrt 2 v{\text{ }}\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j}
Multiplying both sides by $- 1$ we get
\Rightarrow {v_C} - {v_A} = \sqrt 2 v{\text{ }}\left( {{\text{ - }}\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} } \right) …………………..(7)
From (6) and (7) we get
\Rightarrow {v_{CA}} = \sqrt 2 v{\text{ }}\left( {{\text{ - }}\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} } \right)
Thus the relative speed of the ship A with respect to the ship C comes out to be equal to $\sqrt 2 v$ opposite to the direction of the unit vector \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} . According to our assumption, the unit vector \overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} is along the North direction. So the required relative speed is towards the South direction.
Hence, the correct answer is option B.

Note
We should always prefer to solve these types of questions using the unit vectors rather than the method of addition of the vectors. Though this question can also be solved by the latter method, the former method is easier and at the same time chances of committing the mistakes are also less in this method.