Acceleration From Different Frames

Let $\vec { { a }_{ 1 } }$ be the acceleration vector of the particle in frame ${S}_{1}$ and $\vec { { a }_{ 2 } }$ be the acceleration vector of the particle in frame ${S}_{2}$ .

It's given that the magnitude of this acceleration, is same for both the frames, or:

$\left| \vec { { a }_{ 1 } } \right| =\left| \vec { { a }_{ 2 } } \right| =4m/{ s }^{ 2 }$

But, we know that the relative acceleration of the particle in one frame with respect to the other is given by:

$\left| \vec { { a }_{ rel } } \right| =\left| \vec { { a }_{ 2 } } -\vec { { a }_{ 1 } } \right|$

Also, the magnitude of $\vec { { a }_{ 2 } } -\vec { { a }_{ 1 } }$ can be given as:

$\sqrt { { \left| \vec { { a }_{ 1 } } \right| }^{ 2 }+{ \left| \vec { { a }_{ 2 } } \right| }^{ 2 }+2\vec { { a }_{ 1 } } .\vec { { a }_{ 2 } } }$

where $\vec { { a }_{ 1 } } .\vec { { a }_{ 2 } }$ represents the dot product of vectors $\vec { { a }_{ 1 } }$ and $\vec { { a }_{ 2 } }$ . Now, putting values of ${ \left| \vec { { a }_{ 1 } } \right| }$ and ${ \left| \vec { { a }_{ 2 } } \right| }$ , we get ${ \left| \vec { { a }_{ rel } } \right| }$ as:

$\sqrt { 16+16+2.4.4.\cos { \theta } } =\sqrt { 32+32\cos { \theta } }$

where $\theta$ is the angle between the vectors $\vec { { a }_{ 1 } }$ and $\vec { { a }_{ 1 } }$ . Since, the value of $\cos{\theta}$ ranges from $-1$ to $+1$ , so the value of ${ \left| \vec { { a }_{ rel } } \right| }$ ranges from $\sqrt { 32-32 }$ to $\sqrt { 32+32 }$ or $0m/{s}^{2}$ to $8m/{s}^{2}$