The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature R = 2 m. Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is ‘a’. The value of 100a is ______ m/s².

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answer is 8.

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Detailed Solution

Image distance, $v = \frac{uf}{u - f} = \frac{- 24 \cdot 1}{- 24 - 1} = \frac{24}{25}$

Magnification, $m = \frac{- v}{u} = - \frac{24}{25 (- 24)} = \frac{1}{25}$

Mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

Differentiating wrt time gives,

$\frac{- l}{v^{2}} (\frac{dv}{dt}) + \frac{l}{u^{2}} (\frac{du}{dt}) = 0 here, [\frac{dv}{dt} = v_{I}; \frac{du}{dt} = v_{O}]$

$\Rightarrow v_{I} = - (\frac{v^{2}}{u^{2}}) v_{O} = - (\frac{{(\frac{24}{25})}^{2}}{24^{2}}) \times 25 = - \frac{1}{25}$

Again, differentiating wrt time gives,

$\Rightarrow a_{I} = - 2 (\frac{v}{u}) (\frac{v' u - vu'}{u^{2}}) v_{O} = - 2 (\frac{v}{u}) (\frac{v_{I} u - {vv}_{O}}{u^{2}}) v_{O}$

$\Rightarrow a_{I} = - 2 (\frac{\frac{24}{25}}{- \frac{1}{24}}) (\frac{(- \frac{1}{25}) (- 24) - (\frac{24}{25}) (25)}{{(- 24)}^{2}}) (25)$

$\Rightarrow a_{I} = - \frac{2}{25} {ms}^{- 2}$

Thus, $100 |a_{I}| = 100 \times \frac{2}{25} = 8$

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