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Position-time curve of a particle moving along a straight line is shown. Tangents at A and B make angles of $45^{0}$ and $135^{0}$ with positive x-axis respectively. The average acceleration of the particle during $t = 1 s$ to $t = 2 s$ is

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$- 2 m / s^{2}$

$1 m / s^{2}$

$- 1 m / s^{2}$

Zero

answer is A.

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Detailed Solution

$< a > = \frac{V_{B} - V_{A}}{t_{B} - t_{A}} = \frac{\tan 135^{0} - \tan 45^{0}}{2 - 1} = \frac{- 1 - 1}{1} = - 2 m / s^{2}$

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