Question

The accelerations of a particle as seen from two frames $S_1$ and $S_2$ have equal magnitude $4\dfrac{m}{{{s^2}}}$.
(a) The frames must be at rest with respect to each other.
(b) The frames may be moving with respect to each other but neither should be accelerated with respect to the other.
(c) The acceleration of $S_2$ with respect to $S_1$ may either be zero or $8\dfrac{m}{{{s^2}}}$.
(d) The acceleration of $S_2$ with respect to $S_1$may be anything between zero and $8\dfrac{m}{{{s^2}}}$.

Answer 1

A frame of reference is a set of coordinates that can be used to determine positions and velocities of objects in that frame. Another important point is that different frames of reference move relative to one another.

Complete step by step answer:
In this question, we will apply the parallelogram law of vector addition.
According to the parallelogram law of vector addition if two vectors act along two adjacent sides of a parallelogram(whose magnitude equal to the length of the sides) both pointing away from the common vertex, then the resultant is represented by the diagonal of the parallelogram passing through the same common vertex.
According to question,
Let $S_2$ and $S_1$ be the arms of a parallelogram.
So, the formula for the resultant will be,
$R = \sqrt {S_1^2 + S_2^2 + 2{S_1}{S_2}\cos \theta } .......(1)$
Also, in this question we are given that ${S_1} = {S_2}$
We know that the angle between any two vectors can lie between ${0^ \circ }$ and ${180^ \circ }$.
Let us put both the angles in the equation (1),
On putting $\theta = {0^ \circ }$,
$R = \sqrt {S_1^2 + S_2^2 + 2{S_1}{S_2}\cos {0^ \circ }}$
$R = \sqrt {S_1^2 + S_2^2 + 2{S_1}{S_2}}$
Also, it is given that ${S_1} = {S_2} = 4\dfrac{m}{{{s^2}}}$
So,
$R = \sqrt {{4^2} + {4^2} + 2 \times 4 \times 4}$
$R = \sqrt {16 + 16 + 32}$
On further solving,
$R = \sqrt {64}$
$R = 8\dfrac{m}{{{s^2}}}........(2)$
On putting $\theta = {180^ \circ }$,
$R = \sqrt {S_1^2 + S_2^2 + 2{S_1}{S_2}\cos {{180}^ \circ }}$
$R = \sqrt {S_1^2 + S_2^2 - 2{S_1}{S_2}}$
Also, it is given that ${S_1} = {S_2} = 4\dfrac{m}{{{s^2}}}$
So,
$R = \sqrt {{4^2} + {4^2} - 2 \times 4 \times 4}$
$R = \sqrt {16 + 16 - 32}$
On further solving,
$R = \sqrt 0$
$R = 0\dfrac{m}{{{s^2}}}........(3)$
From equation (2) and (3), we can clearly see that the acceleration of $S_2$ with respect to $S_1$may be anything between zero and $8\dfrac{m}{{{s^2}}}$.
So, the correct answer is option (D).

Note: It is important to note that the meaning of rotation and revolution is completely different in context of the solar system. Rotation is the process of moving around anything. Whereas revolution is the process of moving around itself as well as the moving around any other source.