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

1. Key Theory Needed

1. Mirror (Gaussian) Formula
   $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
   where
   • $u$ = distance of object (behind the mirror, so negative by sign convention)
   • $v$ = distance of image (virtual, so positive for convex mirror)
   • $f$ = focal length ($f=\frac{R}{2}$ and is positive for convex)

2. Image Velocity & Acceleration
   Differentiate the mirror equation w.r.t. time $t$. First derivative gives image velocity $\frac{dv}{dt}$. Second derivative gives image acceleration $\frac{d^2v}{dt^2}$.

3. Uniform Object Motion
   If the approaching car moves with constant speed $v_0$, its own acceleration is zero. Hence $\frac{d^2u}{dt^2}=0$.

4. Unit Conversions
   Always change km h$^{-1}$ to m s$^{-1}$: multiply by $\frac{5}{18}$.

2. Logical Chain of Thought

1. Find $f$ from $R$: $f=+1\,\text{m}$ because $R=2\,\text{m}$.
2. Write $u$ at the given instant: $u=-24\,\text{m}$ (negative since object is in front of the reflecting surface).
3. Use mirror formula to get $v$.
4. Differentiate to relate $\frac{dv}{dt}$ to $\frac{du}{dt}$.
5. Differentiate again (or use product-rule carefully) to get $\frac{d^2v}{dt^2}$.
6. Plug numbers: object speed $25\,\text{m s}^{-1}$, $u$, $v$ and simplify. Final numeric value asked is $100a$.

Why each step?

Step 1–3 give the static image position.
Step 4–5 translate real-world motion of the car into motion of its image.
Step 6 brings everything to actual numbers and units.

What is going on?

Imagine you are sitting in a car and holding a curved shaving mirror in your hand. A friend is running towards you from behind. Because the mirror is curved outward (convex), your friend’s reflection is inside the mirror and it is much closer than the real friend.

Now, if your friend keeps running at a steady speed, the reflection seems to move too. Sometimes the reflection even seems to speed up or slow down, even though the real friend is running at the same speed all the time. That ‘speed-up’ or ‘slow-down’ of the reflection is what we call acceleration of the image.

The question is simply asking: “When the real car is 24 m behind, how fast is the reflection’s speed changing?”

Step-by-Step Calculation

1. Focal length
   $f = \frac{R}{2} = \frac{2\,\text{m}}{2} = 1\,\text{m}$

2. Object distance (instant shown)
   $u = -24\,\text{m}$

3. Image distance using mirror formula

   \frac{1}{v} + \frac{1}{u} &= \frac{1}{f} \\ \frac{1}{v} &= \frac{1}{f} - \frac{1}{u} \\ &= 1 - \left(-\frac{1}{24}\right) = 1 + \frac{1}{24} = \frac{25}{24} \\ v &= \frac{24}{25} \;\text{m} = 0.96\,\text{m} \end{aligned}$$

4. Object speed (converted)
   $v_0 = 90\,\text{km h}^{-1} = 90 \times \frac{5}{18} = 25\,\text{m s}^{-1}$ For the object coordinate $u$ (negative),
   $\frac{du}{dt} = +25\,\text{m s}^{-1}$
   (because $u$ becomes less negative as the car approaches).

5. First derivative of mirror formula
   $\frac{d}{dt}\left(\frac{1}{v}+\frac{1}{u}\right)=0 \;\Rightarrow\; -\frac{1}{v^2}\,\frac{dv}{dt} - \frac{1}{u^2}\,\frac{du}{dt}=0$ Hence $\frac{dv}{dt}=\left(\frac{v}{u}\right)^2\!\frac{du}{dt}$

6. Define helper ratio
   $S = \left(\frac{v}{u}\right)^2 = \frac{v^2}{u^2} = \frac{\left(\frac{24}{25}\right)^2}{24^2} = \frac{1}{625}$ So $\frac{dv}{dt}=S\,\frac{du}{dt}=\frac{1}{625}\times25=\frac{1}{25}\,\text{m s}^{-1}$

7. Second derivative (image acceleration)
   With $\frac{d^2u}{dt^2}=0$ (object moves uniformly), $\frac{d^2v}{dt^2}= \frac{d}{dt}\left(S\frac{du}{dt}\right)=\left(\frac{dS}{dt}\right)\frac{du}{dt}$ Using $S=\frac{v^2}{u^2}$, one obtains $\frac{dS}{dt}=\frac{v^2}{u^2}\,2\left(\frac{dv/dt}{v}-\frac{du/dt}{u}\right)$ Substituting in, the algebra simplifies neatly to $a = \frac{d^2v}{dt^2}=2S\left(\frac{du}{dt}\right)^2\frac{(v-u)}{u^2}$

8. Put the numbers

   a &= 2\left(\frac{1}{625}\right)(25)^2\frac{\left(\frac{24}{25}+24\right)}{(24)^2} \\ &= 2(1)(1)\, \frac{24/25 + 24}{576} \\ &= 2\,\frac{624/25}{576} = \frac{1248}{25\times576} \\ &= \frac{26}{25\times12}=\frac{13}{150}\,\text{m s}^{-2} \approx 0.0867\,\text{m s}^{-2} \end{aligned}$$

9. Asked value
   $100a = 100\times0.0867 \approx 8.67\,\text{m s}^{-2}$

Answer: $\boxed{8.67\,\text{m s}^{-2}\text{ (approximately)}}$