91Ó°ÊÓ

The given data can be listed below as:

• The first velocity of the particle is 0 .
• The second velocity of the particle is $Â\pm 1 \mathrm{m}/\mathrm{s}$.
• The third velocity of the particle is$Â\pm 2 \mathrm{m}/\mathrm{s}$ .
• The fourth velocity of the particle is $Â\pm 3 \mathrm{m}/\mathrm{s}$.
• The time taken by the observer is $\mathrm{t}=1 \mathrm{s}.$

Thevelocity is described as the distance a particular object covers in a unit time. The velocity is also directly proportional to the acceleration of an object.

As the particle’s velocities are relative to mainly one reference frame which is say S frame, then the distance will equal the product of the velocity and the time taken. As the time taken is$\mathrm{t}=1\mathrm{s}$ , then the distance will be equal to the velocity such as role="math" localid="1659175841147" $0 \mathrm{m},1 \mathrm{m},2 \mathrm{m}$and $3 \mathrm{m}.$.

Thus, the distance will have the same magnitude as the velocities which are , $0 \mathrm{m},1 \mathrm{m},2 \mathrm{m}$, and 3 m respectively which shows that the velocity is totally dependent on distance. If the time changes, the distance will automatically change.

In a different frame S' which moves with a different velocity, the velocity that is measured in the different frame is the same as the previous velocity. For an observer in the frame , the whole frame moves with the same velocity. Hence, the particle’s velocity in the previous frame S is the same as the new frame and that shows that the magnitude of the relative velocity is the same at different reference frames.

Thus, if there is a different frame, then also the distance will be the same according to the relative velocity. If the time changes, the distance will automatically change.