Q. The velocity of image w.r.t ground in the below figure is

(a) 45 m/s and approaches the mirror

(b) 45 m/s and moves away from the mirror

(c) 60 m/s and approaches the mirror

(d) 60 m/s and moves away from the mirror

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Ans: (a)

Sol: According to mirror formula ,

$ \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $

Differentiating

$ \displaystyle -\frac{dv}{v^2} – \frac{du}{u^2} = 0 $

$ \displaystyle \frac{dv}{du} = -\frac{v^2}{u^2} $

Dividing by dt ,

$ \displaystyle \frac{dv/dt}{du/dt} = -\frac{v^2}{u^2} $

$ \displaystyle \frac{V_I}{V_O} = -(\frac{60}{20})^2 = -9 $

V_{I }= -9V_{o }= -9 × 5 = -45 m/s