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NEET PHYSICSEasy

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

A

1:31:\sqrt{3}

B

3:1\sqrt{3}:1

C

1:11:1

D

1:21:2

Step-by-Step Solution

  1. Principle: The instantaneous velocity of a particle is given by the slope of the tangent to its displacement-time (s−ts-t) graph . v=dsdt=tan⁡θv = \frac{ds}{dt} = \tan \theta where θ\theta 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∘30^\circ. vA=tan⁡30∘=13v_A = \tan 30^\circ = \frac{1}{\sqrt{3}}
  3. Calculate Velocity of B: The angle for particle B is 45∘45^\circ. vB=tan⁡45∘=1v_B = \tan 45^\circ = 1
  4. Calculate Ratio: vAvB=131=13\frac{v_A}{v_B} = \frac{\frac{1}{\sqrt{3}}}{1} = \frac{1}{\sqrt{3}} ratio = 1:31:\sqrt{3}
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