What is the distance between the motion sensor and the cart

[Mr. Goosemahn]

Homework Statement

a.) The ultrasonic motion sensor sends pulses of the ultrasound toward a cart on the low-friction track and determines the distance by the time an echo takes to return. The temperature in the lab is equal to 20 °C. What is the distance between the motion sensor and the cart, if the reflected echo was recorded after 4.17 ms (1 ms = 0.001 s)?

b.) The velocity of sound in air is a function of temperature and increases by 0.6 m/s for every degree Celsius (°C) of temperature increase. The temperature in the lab has increased by = 6.11°C, but students did not noticed that change and they were using the sound velocity value for 20°C (343 m/s). The result of the measurements was incorrect. What is the apparent distance for the increased temperature?

Homework Equations

D=(v*t)/2

The Attempt at a Solution

I got the first answer right, getting 0.715781. I tried using the same equation for the second question, but got nothing. Help?

 

[berkeman]

Homework Statement

a.) The ultrasonic motion sensor sends pulses of the ultrasound toward a cart on the low-friction track and determines the distance by the time an echo takes to return. The temperature in the lab is equal to 20 °C. What is the distance between the motion sensor and the cart, if the reflected echo was recorded after 4.17 ms (1 ms = 0.001 s)?

b.) The velocity of sound in air is a function of temperature and increases by 0.6 m/s for every degree Celsius (°C) of temperature increase. The temperature in the lab has increased by = 6.11°C, but students did not noticed that change and they were using the sound velocity value for 20°C (343 m/s). The result of the measurements was incorrect. What is the apparent distance for the increased temperature?

Homework Equations

D=(v*t)/2

The Attempt at a Solution

I got the first answer right, getting 0.715781. I tried using the same equation for the second question, but got nothing. Help?

Please post the details of your work; we will look for errors. Thanks.

 

[Mr. Goosemahn]

For the first one, I plugged in the velocity (343.3), and the time (4.17 ms, which turns to 0.00417 seconds) and carried out the formula.

D = (343.3*0.00417)/2 = 0.7157805

For the second one, I tried using the same equation and I plugged in 0.6 for V and 26.11 for time, but I got an incorrect answer.

D = (0.6*(6.11/0.6))/2 = 3.05500.

I'm pretty sure I'm doing something wrong regarding the time, but I could be doing everything wrong, quite honestly. I don't know what is the correct equation for this, but that's how I tried solving it.

 

[Integral]

You cannot simply close your eyes and plug numbers into formulas.
Use the numbers given for the temperature change to compute the correct speed of sound for the lab temperaure. Then use the corrected speed of sound in the given equation.