Objects A and B are at the origin. If object A moves to (-6 ,7 )(6,7) and object B moves to (-6 ,-1 )(6,1) over 3 s3s, what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

3 Answers
Apr 29, 2016

8/3ms^-183ms1

Explanation:

Relative velocity is a vector quantity. Lets check component wise:

  • x-component:
    Both the bodies have moved to the same vertical line, namely x= -6x=6 line. Hence their x-components of the velocities are same.

  • y-component:
    The distance between their final positions is: (7-(-1)) = 8" m"(7(1))=8 m
    Hence the relative velocity along y-axis is: 8/3 ms^-183ms1

Hence the relative velocity is 8/3 ms^-183ms1 along y-axis.

Apr 29, 2016

v_A=2,67 " m/s"vA=2,67 m/s

Explanation:

enter image source here
"displacement=8 meters"displacement=8 meters

v_A=("displacement")/("time")vA=displacementtime

v_A=8/3vA=83

v_A=2,67 " m/s"vA=2,67 m/s

Apr 29, 2016

In 3s displacement of A from (0.0) to (-6,7) =vecd_A=(-6hati+7hatj)m=dA=(6ˆi+7ˆj)m
Velocity of A , vecV_A=vecd_A/3=1/3(-6hati+7hatj)m/sVA=dA3=13(6ˆi+7ˆj)ms

In 3s displacement of B from (0.0) to (-6,-1) =vecd_B=(-6hati-hatj)m=dB=(6ˆiˆj)m
Velocity of B, vecV_B=vecd_B/3=1/3(-6hati-hatj)m/sVB=dB3=13(6ˆiˆj)ms
The relative velocity of object B from the perspective of object A
is vecV_B-vecV_A=1/3(-6hati-hatj+6hati-7hatj)=-8/3hatjm/sVBVA=13(6ˆiˆj+6ˆi7ˆj)=83ˆjms
So magnitude of relative velocity is 8/3m/s83ms and it is directed towards negative direction of y-axis i.e. 270^o270o with positive direction of X-axis