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 car in the side view mirror is ' a '. The value of 100a is __________ m/s². Numerical answer type. Work it out, then open the answer.

Answer: 8. Convex mirror: f = +1 m; u = -24 m; du/dt = +25 m/s (90 km/h), constant. v = (fu)/(u - f) ⇒ dv/dt = -(f²)/((u - f)²)du/dt a = (d²v)/(dt²) = (2f²)/((u - f)³)(du/dt)² = (2(1)(625))/((-25)³) |a| = (1250)/(15625) = 0.08 m/s² ⇒ 100a = 8 m/s²

The driver sitting inside a parked car is watching vehicles… | JEE Main 2025 (22 Jan, Shift 1)

Q9JEE Main 2025 · 22 JAN · SHIFT 1Medium

Reflection & Mirrors

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 $\text{R} = 2\text{ m}$ . Another car approaches him from behind with a uniform speed of $90\text{ km/hr}$ . 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 $100a$ is __________ $\text{m/s}^2$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 8

Convex mirror: $f = +1\text{ m}$ ; $u = -24\text{ m}$ ; $\frac{du}{dt} = +25\text{ m/s}$ (90 km/h), constant.
$v = \frac{fu}{u - f} \Rightarrow \frac{dv}{dt} = -\frac{f^2}{(u - f)^2}\frac{du}{dt}$
$a = \frac{d^2v}{dt^2} = \frac{2f^2}{(u - f)^3}\left(\frac{du}{dt}\right)^2 = \frac{2(1)(625)}{(-25)^3}$
$|a| = \frac{1250}{15625} = 0.08\text{ m/s}^2 \Rightarrow 100a = 8\text{ m/s}^2$

Report a mistake

Reflection & Mirrors question from JEE Main 2025 (22 Jan, Shift 1)

More questions on Reflection & Mirrors