The speed of sound, v, in air is a function of the temperature T, of the air…
$v=331.4+0.6(T-273)$ with v in meters per second and T in kelvins. Suppose the rate of change of air temperature is -0.11 K/min. Find the rate of change of the speed of sound with respect to time.
I took the derivative of v and got $v'=0.6T'$. Then I plugged in -0.11 in for T' and got -0.066. I am not sure of the units $m/s^2$ or $m/min^2$ would be my guess but it doesn't seem to work out. Also wondering if the process is right?
Thanks
Best Answer
$$\frac{dv}{dT} = 0.6 \frac{m/s}{K}$$
$$\frac{dT}{dt} = -0.11 \frac{K}{min}$$
Multiplying these together, we get:
$$\frac{dv}{dt} = -0.066 \frac{m/s}{min}$$
The units are weird because it's technically an acceleration: the speed is decreasing by 0.066 meters per second per minute. If you want to write this in SI units, noting that there are 60 seconds in a minute, we can say:
$$\frac{dv}{dt} = -0.0011 \frac{m}{s^2}$$