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 $\mathrm{R}=2 \mathrm{~m}$. Another car approaches him from behind with a uniform speed of $90 \mathrm{~km} / \mathrm{hr}$.…

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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 $\mathrm{R}=2 \mathrm{~m}$. Another car approaches him from behind with a uniform speed of $90 \mathrm{~km} / \mathrm{hr}$. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' a '. The value of 100 a is _______ $\mathrm{m} / \mathrm{s}^2$.

Aicharya · Accepted Answer

Image distance, $v=\frac{u f}{u-f}=\frac{-24 \cdot 1}{-24-1}=\frac{24}{25}$ Magnification, $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=-\frac{24}{25(-24)}=\frac{1}{25}$ Mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ Differentiating wrt time gives, $\begin{aligned} & \frac{-l d v}{v^2} \frac{d t}{d t}+\frac{1}{u^2} \frac{d u}{d t}=0 \text { here, } \frac{d v}{d t}=v_1 ; \frac{d u}{d t}=v_0 \ & \Rightarrow v_1=-\frac{v^2}{u^2} v_0=-\frac{\frac{24^2}{25}}{24^2} \times 25=-\frac{1}{25} \end{aligned}$ Again, differentiating wrt time gives, $\begin{aligned} & \Rightarrow a_{\mathrm{I}}=-2 \frac{v v^{\prime} u-v u^{\prime}}{u^2} v_{\mathrm{O}}=-2 \frac{v v_{\mathrm{I}} u-v v_0}{u^2} v_{\mathrm{O}} \ & \Rightarrow a_{\mathrm{I}}=-2 \frac{\frac{24}{25}-\frac{1}{25}-24-\frac{24}{25} 25}{-24^2} 25 \ & \Rightarrow a_{\mathrm{I}}=-\frac{2}{25} \mathrm{~ms}^{-2} \end{aligned}$ Thus, $100 \mathrm{a}_{\mathrm{I}}=100 \times \frac{2}{25}=8$

The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his…

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