Car B overtakes another car A at a relative speed of 40 ms -1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?

Question

Car B overtakes another car A at a relative speed of 40 ms -1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A? (A) 4 ms -1 (B) 0.2 ms -1 (C) 40 ms -1 (D) 0.1 ms -1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/de165feb7ae1615f7b115d9ff19ecf06-.png" /> Mirror used is convex mirror (rear-view mirror) ∴ V I / m =- m 2 V O / m Given, V O / m =40 m / s m =( f / f - u )=(10/10+190)=(10/200) ∴ V I / m =-(1/400) × 40=-0.1 m / s ∴ Car will appear to move with speed 0.1 m / s.

Q. Car $B$ overtakes another car $A$ at a relative speed of $40 m s^{- 1}$ . How fast will the image of car $B$ appear to move in the mirror of focal length $10 c m$ fitted in car $A$ , when the car $B$ is $1.9 m$ away from the car $A$ ?

$4 m s^{- 1}$

$0.2 m s^{- 1}$

$40 m s^{- 1}$

$0.1 m s^{- 1}$

Mirror used is convex mirror (rear-view mirror)
$∴ V_{I / m} = - m^{2} V_{O / m}$
Given,
$V_{O / m} = 40 m / s$
$m = \frac{f}{f - u} = \frac{10}{10 + 190} = \frac{10}{200}$
$∴ V_{I / m} = - \frac{1}{400} \times 40 = - 0.1 m / s$
$∴$ Car will appear to move with speed $0.1 m / s$ .

Q. Car B overtakes another car A at a relative speed of 40ms−1. How fast will the image of car B appear to move in the mirror of focal length 10cm fitted in car A, when the car B is 1.9m away from the car A?

4ms−1

0.2ms−1

40ms−1

0.1ms−1