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.

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Solution

Q.
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 ________ $m / s^{2}$ . [2025]

Ans.
(8)

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

$\frac{1}{v} + (\frac{1}{- 24}) = 1 \Rightarrow v = \frac{24}{25} cm$

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

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ ... (i)

Diff. (i) w.r.t. time

$\frac{- 1}{v^{2}} (\frac{d v}{d t}) - \frac{1}{u^{2}} (\frac{d u}{d t}) = 0$ ... (ii)

$V_{1} = - m^{2} V_{0} = - {(\frac{1}{25})}^{2} \cdot 25 = - \frac{1}{25} {ms}^{- 1}$

Diff. (ii) w.r.t. time

$- (- \frac{2}{v^{3}} \cdot {(V_{I})}^{2} + \frac{1}{v^{2}} \cdot a_{I}) - (\frac{2}{u^{3}} \cdot {(V_{O})}^{2} + \frac{1}{u^{2}} \cdot a_{O}) = 0$

$\frac{a_{I}}{v^{2}} = \frac{2}{v^{3}} \cdot V_{I}^{2} - \frac{2}{u^{3}} \cdot V_{O}^{2}$

$\Rightarrow a_{I} = \frac{2}{v} \cdot V_{I}^{2} - \frac{2 v^{2}}{u^{3}} \cdot V_{O}^{2}$

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

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

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