Sudmarine sonar/distance problem

  • Thread starter Thread starter brick
  • Start date Start date
AI Thread Summary
The discussion revolves around a submarine's sonar ping and the subsequent echo it receives from a battleship. The submarine, traveling at 10 m/s, hears an echo 9 seconds after the ping, while the echo takes 8 seconds to return. The sound travels in salt water at 1533 m/s, leading to a calculated distance of 12,264 meters. The confusion arises regarding the battleship's speed, as initial calculations suggest it is also traveling at 10 m/s, which seems incorrect given both vessels are in motion. The problem requires a reevaluation of the distances covered by both vessels during the ping's travel time to accurately determine the battleship's speed and the time until the submarine is directly below it.
brick
Messages
3
Reaction score
0

Homework Statement


A submarine pings its sonar and 9 seconds later hears an echo, the signature of which indicates a battleship. Ten seconds later, the ping takes 8 seconds to echo back. The submarine is traveling at 10 m/s towards the battleship. The battleship is directly in front of the submarine, traveling at the same bearing, at an unknown speed. The salt water is at a temp of 20 degrees C. How long until the sub is directly below the battleship, assuming no change in course or speed in either vessel? What is the speed of the battleship?

Homework Equations


Sound travels in the salt water at 1533 m/s at 20 degrees C.
V= d/t
I am sure there are more but that is what i used

The Attempt at a Solution


1.) v=d/t
1533 = d/8
d = 1533 * 8
d = 12264 m

2.)
v = d/t

10 m/s = 12264 m/t
t = 12264 / 10
t = 1226.4 seconds time for the sub to reach the battleship

3.) v=d/t
v = 12,264 m / 1226.4 s

v= 10 m/s for the battleship speed which this is where i am getting hung up on i don't believe this is right cause it has the same velocity as the sub. Any help would be appreciated thank you.
 
Physics news on Phys.org
Keep in mind that both ships are in motion during the time the ping emits to when it is detected.

Hence the distance the ping travels is the distance between the 2 plus the distance the battleship moved until pinged and then the distance covered back to the sub by the echo is diminished by the distance the sub has traveled until reception.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top