Q. The accelerations of a particle as observed from two different frames $S_{1}$ and $S_{2}$ have equal magnitudes of $2\, ms ^{-2}$.

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Solution:

Acceleration of particle w.r.t. frame $S_{1}: \vec{a}_{p}-\vec{a}_{s_{1}}=2 \hat{n}$
Acceleration of particle w.r.t. frame $S_{2}: \vec{a}_{p}-\vec{a}_{s_{2}}=2 \hat{m}$
where $\hat{m}$ and $\hat{n}$ are unit vectors in any directions. Now relative acceleration of frames:
$\vec{a}_{s_{2}}-\vec{a}_{s_{1}}=2(\hat{n}-\hat{m})$
Its magnitude can have any value between 0 to $4\, ms ^{-2}$, depending upon the directions of $\hat{m}$ and $\hat{n}$.