The displacement-time (s-t) graphs of two moving particles A and B make angles of $30^\circ$ and $45^\circ$ wi

Question

The displacement-time (s-t) graphs of two moving particles A and B make angles of $30^\circ$ and $45^\circ$ with the x-axis as shown in the figure. The ratio of their respective velocity $(v_A/v_B)$ is:

Accepted Answer

1.  **Principle:** The instantaneous velocity of a particle is given by the slope of the tangent to its displacement-time ($s-t$) graph .
    $$v = \frac{ds}{dt} = 	an 	heta$$
    where $	heta$ is the angle made by the graph with the time axis (x-axis).
2.  **Calculate Velocity of A:** The angle for particle A is $30^\circ$.
    $$v_A = 	an 30^\circ = \frac{1}{\sqrt{3}}$$
3.  **Calculate Velocity of B:** The angle for particle B is $45^\circ$.
    $$v_B = 	an 45^\circ = 1$$
4.  **Calculate Ratio:**
    $$\frac{v_A}{v_B} = \frac{\frac{1}{\sqrt{3}}}{1} = \frac{1}{\sqrt{3}}$$
    ratio = $1:\sqrt{3}$

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