Verified Solution • 2025–26 Session moderate

QJEEM25Ph26bb: JEE Main 2025 Physics - Class 12 JEEM25Ph26bb Solution: The driver sitting inside a parked car is watching vehicles…

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\mathrm{R}=2 \mathrm{~m} . Another car approaches him from behind with a uniform speed of 90 km/hr90 \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 ' aa '. The value of 100a100 a is __________ m/s2\mathrm{m} / \mathrm{s}^2 .

Step-by-Step Verified Solution

amp;amp;v=ufu−f=−24⋅1−24−1=2425amp; m=−vu=−2425(−24)=125amp;1v−1u=1famp;vI=−m2⋅v0=−1(25)2⋅25=−125amp; Diff. \begin{aligned}&\begin{aligned}& \mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}}=\frac{-24 \cdot 1}{-24-1}=\frac{24}{25} \\& \mathrm{~m}=\frac{-\mathrm{v}}{\mathrm{u}}=-\frac{24}{25(-24)}=\frac{1}{25} \\& \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\& \mathrm{v}_{\mathrm{I}}=-\mathrm{m}^2 \cdot \mathrm{v}_0=\frac{-1}{(25)^2} \cdot 25=\frac{-1}{25}\end{aligned}\\&\text { Diff. }\end{aligned}

amp;−1v2(dvdt)+1u2(dudt)=0[dvdt=vI;dudt=v0]amp;+2v3⋅(vI)2−1v2⋅aI−2u3⋅(v0)2+1u2⋅a0=0amp;aIv2=2v3⋅vI2−2u3⋅v02amp;aI=2v⋅vI2−2v2u3⋅v02amp;=2⋅2524⋅125⋅125−2(24)3⋅2425⋅2425⋅25⋅25amp;aI=224.25−224amp;aI=224⋅−2425=−225amp;100aI=225×100=8\begin{aligned}& \frac{-1}{\mathrm{v}^2}\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)+\frac{1}{\mathrm{u}^2}\left(\frac{\mathrm{du}}{\mathrm{dt}}\right)=0 \quad\left[\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{I}} ; \frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{v}_0\right] \\& \frac{+2}{\mathrm{v}^3} \cdot\left(\mathrm{v}_{\mathrm{I}}\right)^2-\frac{1}{\mathrm{v}^2} \cdot \mathrm{a}_{\mathrm{I}}-\frac{2}{\mathrm{u}^3} \cdot\left(\mathrm{v}_0\right)^2+\frac{1}{\mathrm{u}^2} \cdot \mathrm{a}_0=0 \\& \frac{\mathrm{a}_{\mathrm{I}}}{\mathrm{v}^2}=\frac{2}{\mathrm{v}^3} \cdot \mathrm{v}_{\mathrm{I}}^2-\frac{2}{\mathrm{u}^3} \cdot \mathrm{v}_0^2 \\& \mathrm{a}_{\mathrm{I}}=\frac{2}{\mathrm{v}} \cdot \mathrm{v}_{\mathrm{I}}^2-\frac{2 \mathrm{v}^2}{\mathrm{u}^3} \cdot \mathrm{v}_0^2 \\& =\frac{2 \cdot 25}{24} \cdot \frac{1}{25} \cdot \frac{1}{25}-\frac{2}{(24)^3} \cdot \frac{24}{25} \cdot \frac{24}{25} \cdot 25 \cdot 25 \\& \mathrm{a}_{\mathrm{I}}=\frac{2}{24.25}-\frac{2}{24} \\& \mathrm{a}_{\mathrm{I}}=\frac{2}{24} \cdot \frac{-24}{25}=\frac{-2}{25} \\& 100 \mathrm{a}_{\mathrm{I}}=\frac{2}{25} \times 100=8\end{aligned}

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