Question

a road runs parallel to a vertical Cliff at a distance of 495 M as shown in the figure A car standing at A blow the horn and driver of the car hear the Echo after 3 seconds but a person standing at B hear the sound of horn twice within an interval of 2 seconds find the distance between car and the person.

Answer 1

Answer:

Explanation:

Firstly find the velocity using data given for person A and then use same speed for person B but there are 2 ways for sound

Answer 2

Distance between the car and the person is 330 m.

Explanation:

Given:

           distance between the driver A and the cliff 'd' = 495 m

           Driver hears echo after 't' = 3 sec.

           Person standing near to cliff hear sound after 'T' = 1 sec.

Person hears two echoes in 2 second which means first he heard the original sound the  the echo sound in 2 seconds.

So,

                 ⇒  Speed 's' = $\frac{disyance }{time}$

                 ⇒ s = $\frac{d\times2}{t}$

         'd x 2' is taken because sound first hit the cliff then return back. It travels twice the distance.

         

              ⇒ s = $\frac{ 495 \times 2}{3}$

               ⇒ s = 165 x 2 m/s

As we know, the speed of sound will remain same for both of them.

              Now from the question,

                         for person at b position.

                        ⇒ 165 x 2 = $\frac{495 + x}{2}$

where 'x' is the distance of person B from the cliff.

               

                         ⇒ x = 165 m

      Distance between the driver and person at point B is

                    = d - x

                    = 495 - 165

                    = 330 m