Question

The displacement time graphs of two particles A and B are straight lines making angles of 30° and 60° respectively with the time axis. If the velocity of A is \[{v_A}\]​ and that of B is \[{v_B}\] ​, then the value of \[\dfrac{{{v_A}}}{{{v_B}}}\]​​ is
A. \[\dfrac{1}{2}\]
B. \[\dfrac{1}{{\sqrt 3 }}\]
C. \[\sqrt 3 \]
D. \[\dfrac{1}{3}\]

Answer 1

Hint:The slope of the displacement-time graph of a body is the rate change of displacement with respect to time. As we know that the rate of change of displacement is equal to the velocity of the body. Hence, the velocity of the body is the slope of the displacement-time graph.

Formula used:
\[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
Here, m is the slope of the line joining two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\].
\[m = \tan \theta \]
Here m is the slope of the line making an angle of \[\theta \] with the horizontal.

Complete step by step solution:
For body A, the slope of the displacement-time graph is constant and the line is making an angle of 30° with the axis which represents time. So, the velocity of the body A will be the tangent of angle 30° It is given that the velocity of the body A is \[{v_A}\], then
\[{v_A} = \tan 30^\circ \]
\[\Rightarrow {v_A} = \dfrac{1}{{\sqrt 3 }}m{s^{ - 1}}\]

For body B, the slope of the displacement-time graph is constant and the line is making an angle of 60° with the axis which represents time. So, the velocity of the body B will be the tangent of angle 60°. It is given that the velocity of the body B is \[{v_B}\], then
\[{v_B} = \tan 60^\circ \]
\[\Rightarrow {v_B} = \sqrt 3 m{s^{ - 1}}\]

We need to find the ratio of the velocity of the body A to the velocity of the body B,
\[\dfrac{{{v_A}}}{{{v_B}}} = \dfrac{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)m{s^{ - 1}}}}{{\sqrt 3 m{s^{ - 1}}}}\]
\[\Rightarrow \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{1}{{\sqrt 3 \times \sqrt 3 }}\]
\[\therefore \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{1}{3}\]

Therefore, the correct option is D.

Note: A ratio is a pure number, meaning it has no dimensions. The vector quantity known as velocity has both magnitude and direction. We use the acute angle formed by the horizontal axis and the displacement-time graph to determine the magnitude.