How Do You Calculate Distance and Velocity Using Echo Times?

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Homework Help Overview

The problem involves a motion detector that measures the time taken for sound echoes to return after reflecting off a moving car. The scenario includes two clicks emitted by the detector, with the times for the echoes being recorded. The main questions are about calculating the distance the car has moved and determining its average velocity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations based on the echo times and the speed of sound. There is a suggestion to consider the Doppler Effect, questioning whether it is relevant to the scenario. Another participant proposes a coordinate system approach to solve the problem, introducing geometric considerations and distance formulas.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants are providing guidance on how to approach the problem, while others are questioning the assumptions made, such as the relevance of the Doppler Effect. There is no explicit consensus yet on the method to be used.

Contextual Notes

Participants note that the problem is part of a basic lab project focused on displacement and velocity, and that the Doppler Effect has not been introduced in their studies. The speed of sound is confirmed to be 340 m/s, which is a critical piece of information for the calculations.

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Homework Statement



A motion detector transmits a click which is reflected off of a moving car. The echo is received by the motion detector after .0026 seconds. the motion detector sends out a second click .1 seconds after the first click. The echo for the seconds click is received after .0031 seconds.

1) How far has the car moved?
2) What is the average velocity?

Homework Equations



v = Vo + at
S = 1/2 (Vo+V)t
U = s/t
S = vt

The Attempt at a Solution


.0026 - .1 = .0974 seconds
.0031
.0974 + .0031 = .1005 seconds
340 (Spd of sound m/s) * .1005 = 34.17 meters ??
Velocity ??
 
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Are you sure that Doppler Effect does not take place Here??
 
My son and I have posted this question because we have spent all day trying to solve it. The question comes from a basic lab project that deals with displacement and velocity. doppler has not been introduced. I have typed in all the info from the question. The only additional piece to add is that the speed of sound is 340 m/s.
 
Your question can be easily solved by assuming coordinate system(I hope that coordinate axis have been introduced). Assume motion detector lying along X-axis and signal transmitter lying at origin(0,0). Assume car traveling along line y=1(see attachment). Take 2 positions along this line as in your question.Let 1 st position(posn in attachment) be of coordinates (x,1) & second position be (y,1). Calculate distance between 1 and origin using distance formula.Let this be m.
Now (2m/t1)=v
Where t1 is time for first echo(given) and v=velocity of sound.
Similarly calculate distance between second point and origin and let this be n.
Now (2n/t2)=v
Where t2 is time for second echo(given) and v=velocity of sound.

Using 2 equations above calculate x and y individually and again apply distance formula between points (x,1) and (y,1) where we have now found out x & y.

PS:I am no great hands at Drawing and i accept this diagram is filthy.But i hope you understand.
 

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