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The distance-time graph of a particle at time ‘t’ makes angle $45 °$ with the time axis. After one second, it makes angle $60 °$ with the time axis. The average acceleration of the particle is

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$\sqrt{3} - 1$

$\sqrt{3} + 1$

$\sqrt{3}$

$1$

answer is A.

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Detailed Solution

$a = \frac{v - u}{t} (t = 1 s e c)$
$u = \frac{d s}{d t} = \tan 45 ° = 1$
$v = \tan 60^{\circ} = \sqrt{3}$

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