Question

*Nice Question*

A man is riding a vehicle towards a vertical cliff with a velocity of 36 km/h.
He fires a gunshot and hears its echo after 4 sec. The speed of sound in
air is 320 m/s.

(i) The distance between the cliff and the position, where he shot
the gun is:
A. 640 m
B. 680 m
C. 620 m
D. 660 m

(ii) If the speed of the vehicle increases by 4 times, then the time
taken to hear the echo is:
A. 11/3 seconds
B. 10/3 seconds
C. 3 seconds
D. None of the above

Answer 1

Answer:

Correct option is

D

310ms−1

The speed of the sound is calculated as follows.

Let d be the distance between the man and the hill in the beginning.

v=t2×d = 52×d→eqn1

He moves 310 m towards the hill. Therefore the distance will be (d - 310) m. Therefore,

v=32(d−310)→eqn2

Since velocity of sound is same, equating (1) and (2), we get

52d=32(d−310)

3d = 5d - 1550

2d = 1550

d = 775 m

Hence, the velocity of sound v=52×775 (substituting in equation 1)

v=310m/s

Answer 2

Answer:

A man is riding a vehicle towards a cliff with a velocity $36km/h$, he fires a gunshot and hears its echo after $4s$, if the speed of the sound in air is $320m/s$, the distance between the cliff and the position of the vehicle is $660m$. Now when the velocity of the vehicle increased by $4$ times then the time is taken to hear an echo is $11/3s$

Explanation:

Let the distance between the cliff and the position where he shot the gun be $d$.

As he fires a gunshot while riding a vehicle the speed of the vehicle added to the speed of the sound.

Assume that $t_{1}$ is the time taken by the sound to travel from the position of the vehicle to the cliff and $t_{2}$ is the time taken to the reverse travel.

Now, $t_{1} =d/(v_{1}+v_{2})$

and $t_{2} =d/(v_{1}+v_{2} )$

From the question, we have $t_{1}+t_{2} =4s$ , since he hears its echo after $4s$

Adding the equations $1$&$2$ we get

$t_{1} +t_{2}=2d/(v_{1} +v_{2} )$

velocity of vehicle $v_{1} =36km/h=10m/s$

the velocity of sound in air $v_{2}=320m/s$

after rearranging the terms to get distance,

$d=4(330)/2=660m$

If the speed of the vehicle increased by $4$ times, the velocity of the vehicle will be $v_{1}=40m/s$

Then, $t_{1} +t_{2} =2d/(v_{1} +v_{2} )=2(660)/360=11/3s$