Objects A and B are at the origin. If object A moves to (3 ,6 )(3,6) and object B moves to (6 ,-5 )(6,5) over 1 s1s, what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

1 Answer
Jul 30, 2017

vec v_(B w.r.t A)=3 hat i - 11 hat jvBw.r.tA=3ˆi11ˆj

Explanation:

vec r_iri= initial displacement vector
vec r_jrj=final displacement vector
vec r_ArA=Displacement vector of A
vec r_BrB=Displacement vector of B
w.r.t=with respect to

vec r_i=vec 0ri=0
vec r_f=3hati + 6hatjrf=3ˆi+6ˆj
vec r_A= vec r_f - vec r_irA=rfri
vec r_A=3hati + 6hatjrA=3ˆi+6ˆj

vec r_i=vec 0ri=0
vec r_f=6hati - 5hatjrf=6ˆi5ˆj
vec r_B= vec r_f - vec r_irB=rfri
vec r_B=6hati - 5hatjrB=6ˆi5ˆj

vec r_(B w.r.t A) = vec r_B - vec r_ArBw.r.tA=rBrA

vec r_(B w.r.t A) = 3hati - 11hatj (m)rBw.r.tA=3ˆi11ˆj(m)

vec v_(B w.r.t A)=(vec r_(B w.r.t A))/tvBw.r.tA=rBw.r.tAt

vec v_(B w.r.t A)= (3hati - 11hatj (m))/(1s)vBw.r.tA=3ˆi11ˆj(m)1s

vec v_(B w.r.t A)= 3hati - 11hatj (m*s^-1) vBw.r.tA=3ˆi11ˆj(ms1)

|vec v_(Bw.r.tA)| = sqrt(130) m*s^-1vBw.r.tA=130ms1