In the situation as shown in Figure, `(cos53^(@)=(3)/(5))`

Options

A. velocity of image wrt mirror is `-22hat(i)-24hat(j)`
B. velocity of image wrt mirror is `-44hat(i)-24hat(j)`
C. velocity of image wrt ground is `-46hat(i)-24hat(j)`
D. velocity of image wrt ground is `-24hat(i)-24hat(j)`

Answer

b.,c.

Explanation

`vecv_(0)=` velocity of object `=(9hat(i)+12hat(j))ms^(-1)`
`vecv_(m)=` velocity of mirror`=-2hat(i)ms^(-1)`
`m+(f)/(f-u)=(-20)/(-20-(-30))=-2`
For velocity component parallel to optical axis,
`(vecv_(I//m))_(||)=-m^(2)(vecv_(O//m))_(||)`
`(vecv_(I//m))_(||)=-(-2)^(2)11hat(i)=-44hat(i)ms^(-1)`
For velocity component perpendicular to optical axis
`(vecv_(I//m))_(_|_)=m(vecv_(o//m))_(_|_)=-(-2)12hat(j)=-24hat(j)ms^(-1)`
`(vecv_(I//m))=` velocity of image with respet to mirror
`vecv_(I//m)=(vecv_(I//m))_(||)+(vecv_(I//m))_(_|_)=(-44hat(i)-24hat(j))ms^(-1)`
Also, `vecv_(I//m)=vecv_(I)-vecv_(m)`
or `vecv_(I)=vecv_(I//m)+vecv_(m)=(-44hat(i)-24hat(j))-2hat(i)=(-46hat(i)-24hat(j))ms^(-1)`