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 →vS′S=(1.0ˆi+2.0ˆj+3.0ˆk)tm/s. (a) If at time t = 0 the origins coincide, what is the position of origin O′ in the S frame as a function of time?

a) →r(t) = (tˆi + 2tˆj + 3tˆk)m
b) →r(t) = (tˆi + 2tˆj + 3tˆk)m/s
c) →r(t) = (tˆi + 2tˆj + 3tˆk)m/s²
d) →r(t) = (tˆi + 2tˆj + 3tˆk)km/s

Answer 1

The correct answer is (a) →r(t) = (t→i + 2t→j + 3t→k)m, representing the position of origin O' in frame S as a function of time when S' moves away at a constant velocity.

Step-by-step explanation:

If at time t = 0 the origins coincide and the reference frame S' moves away from S at a constant velocity →vS'S=(1.0→i+2.0→j+3.0→k) m/s, then the position of origin O' in the S frame as a function of time t can be found using the equation for position in uniformly moving systems, →r(t) = →v*t. This implies that →r(t) will be (1.0*t→i + 2.0*t→j + 3.0*t→k) meters, where t is the time in seconds. Therefore, the correct answer is (a) →r(t) = (t→i + 2t→j + 3t→k)m.