Let the point of observation be A. 3 minutes is the time for the angle of depression to change from 30 ^\circ to 60^\circ .

So, if the speed of car = v metre/minute, and the time taken to reach from D from B is t minutes.

Then distance between BD = vt and distance between DC = 3v.

\angle ACB= 30^\circ

\angle ADB= 60^\circ

In \Delta ABC

tan 30^\circ = \frac{AB}{BC}

\frac{1}{\sqrt{3}} = \frac{AB}{vt+3v} (since BC = vt - 3v)

AB = \frac{ vt+3v}{\sqrt{3}} equation 1

In \Delta ABD

Tan 60^\circ = \frac{AB}{BD}

\sqrt{3} = \frac{AB}{vt}

\sqrt{3}vt = AB

\sqrt{3}vt = \frac{3v+vt}{\sqrt{3}} (From equation 1)

3v+vt = 3vt

3v = 2vt

2t = 3 (cancelling v on either side)

t = \frac{3}{2}= 1.5 minutes