Question

The coordinate axes of the reference frame S′ remain parallel to those of S, as S′ moves away from S at a constant velocity →vSS′=(4.0ˆi+3.0ˆj+5.0ˆk)m/s. (a) If at time t = 0 the origins coincide, what is the position of the origin O′ in the S frame as a function of time?

a) →r(t) = (4tˆi + 3tˆj + 5tˆk)m
b) →r(t) = (4tˆi + 3tˆj + 5tˆk)m/s
c) →r(t) = (4tˆi + 3tˆj + 5tˆk)m/s²
d) →r(t) = (4tˆi + 3tˆj + 5tˆk)km/s

Answer 1

The position of the origin O' of reference frame S' in reference frame S, as it moves away at a constant velocity, is described by the position vector →r(t) = (4tˆi + 3tˆj + 5tˆk)m.

Step-by-step explanation:

The student's question revolves around basic concepts of kinematics in classical mechanics, specifically within the context of different reference frames. When reference frame S' moves away from reference frame S at a constant velocity, the position of the origin O' in the S frame as a function of time can be described by the equation:

→r(t) = (4tˆi + 3tˆj + 5tˆk)m

This follows from integrating the constant velocity vector over time, where the components of the velocity are multiplied by the time to provide the displacements along the respective axes. The position function is linear with respect to time, indicating there is no acceleration involved and that speed is constant.