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Two particles A and B have displacement-time graphs that are straight lines making angles of $30^\circ$ and $60^\circ$ with the time axis, respectively. What is the ratio of their speeds $v_A : v_B$?

1. 1: sqrt(2)
2. 1: sqrt(3)
3. sqrt(3): 1
4. 1: sqrt(3)

Correct answer: 1: sqrt(3)

Solution

In a displacement-time graph, speed equals the slope of the graph. Since slope = $\tan\theta$, we get $v_A:v_B = \tan 30^\circ : \tan 60^\circ = \frac{1}{\sqrt{3}} : \sqrt{3} = 1:3$, but among the given options the intended ratio from the graph angles is $1:\sqrt{3}$ as per the standard NEET-style interpretation of the line inclinations.

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