Question

Ram moves in east direction at a speed of $6\,m{s^{ - 1}}$ and Shyam moves $30\,{}^ \circ $ east of north at a speed of $6\,m{s^{ - 1}}$. The magnitude of their relative velocity is:
(A) $3\,m{s^{ - 1}}$
(B) $6\,m{s^{ - 1}}$
(C) $6\sqrt 3 \,m{s^{ - 1}}$
(D) $6\sqrt 2 \,m{s^{ - 1}}$

Answer 1

Hint The relative velocity is the difference of the velocity of the one person to another. The velocity of the ram is given directly in the question. The velocity of the Shyam is determined by creating the vector equation, so that the velocity of the Shyam is determined. Then the magnitude of the difference of the velocity vector equation of the two velocities is the relative velocity.

Complete step by step solution
Given that,
The velocity of the Ram is given as, ${V_R} = 6\,m{s^{ - 1}}$,
The velocity vector equation of the ram is given by, ${\vec V_R} = 6\hat i$,
The Shyam moves $30\,{}^ \circ $ east of north at a speed of $6\,m{s^{ - 1}}$.
From the given information, the free body diagram is given as,

From the above diagram, the velocity vector equation of the Shyam is given by,
${\vec V_S} = 6\sin {30^ \circ }\hat i + 6\cos {30^ \circ }\hat j$
Now, the relative velocity is,
${\vec V_{RS}} = {\vec V_R} - {\vec V_S}$
By substituting the velocity vector equation of the Ram and Shyam in the above equation, then
${\vec V_{RS}} = 6\hat i - 6\sin {30^ \circ }\hat i - 6\cos {30^ \circ }\hat j$
The value of the $\sin {30^ \circ } = \dfrac{1}{2}$ and \[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\], then the above equation is written as,
\[{\vec V_{RS}} = 6\hat i - 6 \times \dfrac{1}{2}\hat i - 6 \times \dfrac{{\sqrt 3 }}{2}\hat j\]
By dividing the terms in the above equation, then the above equation is written as,
\[{\vec V_{RS}} = 6\hat i - 3\hat i - 3\sqrt 3 \hat j\]
By subtracting the terms in the above equation, then the above equation is written as,
\[{\vec V_{RS}} = 3\hat i - 3\sqrt 3 \hat j\]
By taking modulus on both sides, then the above equation is written as,
\[\left| {{{\vec V}_{RS}}} \right| = \sqrt {{3^2} + {{\left( {3\sqrt 3 } \right)}^2}} \]
By taking square in the RHS, then the above equation is written as,
\[\left| {{{\vec V}_{RS}}} \right| = \sqrt {9 + 27} \]
By adding the terms in the above equation, then the above equation is written as,
\[\left| {{{\vec V}_{RS}}} \right| = \sqrt {36} \]
By taking the square root in the above equation, then
\[\left| {{{\vec V}_{RS}}} \right| = 6\,m{s^{ - 1}}\]

Hence, the option (B) is the correct answer.

Note While solving the two vector equations the coefficient of the $\hat i$ of the first equation is added, subtracted or multiplied with the coefficient of the $\hat i$ of the second equation. In the velocity vector equation of the Shyam, the sine component is in the $X$ axis, so the sine component is written as the coefficient of the $\hat i$.