Question

A car running with a velocity of 30m/s reaches midway between two vertical parallels walls seperated by 360 m, when driver sounds the horn the first three echoes will be heard by the driver respectively at

Answer 1

#BAL

Sol. x = 30t

For first echo,

  1

1

180 180 – x t

330

360 – 30t1 = 330t1

1   360 t 1s

For second echo,

  2

2

720 x t

330

720 + 30t2 = 330t2

2   720 t 2.4 s

300

For third echo,

 3

3

1080 – x t

330

1080 – 30t3 = 330t3

t3 = 3 s

Answer 2

Answer:

t1,t2,t3 = 1sec, 2sec, 3sec

Explanation:

Let the first echo heard after t 1

​second:in t 1

​ second, distance traveled by the car =v∗t 1 =30t 1

​distance traveled by sound for first echo=d−30t 1

​vs∗t 1 =360−30t 1

​330∗t 1 =360−30t1

​360t1 =360​

t1 =1sec

Let the second echo heard after t2 second:in t2

​second, distance traveled by the car =v∗t2=30t 2

​distance traveled by sound for second echo=2d−30t2

​vs∗t2 =2∗360−30t2 (here vs is velocity of the sound 330m/s)

​330∗t2 =720−30t2

​360t2 =720

t2 ​=2sec

Let the third echo heard after t3 second in t3

​second, distance traveled by the car =v∗t3 =30t3

​distance traveled by sound for third echo=3d−30t 3

​vs∗t3 =3∗360−30t3

​330∗t3 =1080−30t3

​360t3 =1080

t3 = 3sec

t1, t2, t3=1sec,2sec,3sec