Question

To the captain of a ship $A$, which is travelling with velocity ${\vec v_A} = (3\hat i - 4\hat j)\dfrac{{km}}{h}$, a second ship $B$ appears to have a velocity of $(5\hat i + 12\hat j)\dfrac{{Km}}{h}$ . What is the true velocity of the ship $B$?

Answer 1

As we know that, Relative velocity is defined as the velocity of one body with respect to another body, for example is a body has a velocity $V_1$ and another body has velocity $V_2$ then velocity of body one with respect to another is calculated as $V_1 - V_2$.

Complete step by step answer:
Here we know that a two dimensional vector is represented by their two mutually perpendicular unit vectors which are denoted as $\hat i$,$\hat j$ . So, it’s clear that we have given the velocities vectors in two dimensional. We will use relative velocity concept to find actual velocity of ship $B$.

Given, the true velocity of ship A is ${\vec v_A} = (3\hat i - 4\hat j)\dfrac{{km}}{h}$
And ship $B$ appears to have a velocity of $(5\hat i + 12\hat j)\dfrac{{Km}}{h}$ which can be understand as, the relative velocity of ship B with respect to ship A is given by ${\vec v_{BA}} = {\vec v_B} - {\vec v_A}$, where, ${\vec v_B}$ is the true velocity of ship $B$.
On Putting the values of velocity of ship A and relative velocity of ship B with respect to ship A $({\vec v_{BA}})$ on above equation,
We get, $5\hat i + 12\hat j = {\vec v_B} - 3\vec i + 4\vec j$
${\vec v_B} = (8\hat i + 8\hat j)\dfrac{{Km}}{h}$ .

Hence, the true velocity of ship B is ${\vec v_B} = (8\hat i + 8\hat j)\dfrac{{Km}}{h}$.

Note: Remember when two bodies moving in same direction their relative velocities will get subtracted which is also called velocity of approach and when two bodies move in opposite direction, relative velocities will get added and it’s called velocity of separation.