A car is moving with a constant speed of `60 km h^-1` on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of `100 m` and is approaching with a speed of `5 km h^-1`. In order to keep track of the car in the rear, the driver begins to glane alternatively at the rear and side mirror of his car after every `2 s` till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct ?

Options

A. The speed of the car in the rear is `"65 km h"^(-1)`
B. In the side mirror the car in the rear would appear to approach with a speed of `"5 km h"^(-1)` to the driver of the leading car.
C. In the rear view mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases.
D. In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

Answer

D

Explanation

(a) The speed of the car in rear view mirror cannot be `65 km//h` as the mirror is convex. So, this option is incorrect.
(b) The curvature of the rear mirror can be different than that of the side mirror. So, this option is incorrect.
(c) Using mirror.s formula,
`(1)/(f)=(1)/(v)+(1)/(u)`
Differentiating both sides w.r.t. to t and arranging, we get
`-(1)/(v^(2))(dv)/(dt)=(1)/(u^(2))(du)/(dt)`
`rArr" "(dv)/(dt)=(v^(2))/(u^(2))((du)/(dt))`
`"Also, "v=(fu)/(u-f)`
`therefore" "(v^(2))/(u^(2))=((f)/(u-f))^(2)`
`rArr" "(dv)/(dt)=-(f)/(u-f)^(2)((du)/(dt))`
As the distance between the cars decreases, the speed of the image in the side view mirror increases. The correct option is (d).