# A Modest Temperature Gradient

A supersonic aircraft generates a shock wave. At the wave's origin, the temperature is measured to be at $$300\text{ K}$$, but in its direction of travel, the temperature gradient can be modeled by the expression $$\left( -\frac { 2 }{ 75 } d \right)\text{ K}\cdot \text{ km}^{ -2 }$$, where $$d$$ is the distance from the wave's origin in kilometers. (Note: For this expression, plug in the value and the units for $$d$$.)

Using the given approximations, calculate the time (in seconds) that it will take an observer $$15 \text{ km}$$ away from the aircraft to hear the shock wave after it is created. If this can be written in the form $$a\sqrt { b }$$ seconds, where $$a$$ and $$b$$ are positive integers with $$b$$ square-free, write your answer in the form $$a+b$$.

Details and Assumptions

• $$\arcsin { \theta \approx \theta }$$ for small values of $$\theta$$.
• Use the following approximation: $$\sqrt { \dfrac { { { M }_{ a } } }{ { \gamma }_{ a }R } } \approx \frac { 1 }{ 20 }$$.
• $${ M }_{ a }\approx 2.897\times { 10 }^{ -2 }\text{ kg}\cdot \text{ mol}^{ -1 }$$, molar mass of dry air.
• $${ \gamma }_{ a }\approx 1.400$$, adiabatic index/heat capacity ratio of dry air at $$293\text{ K}$$.
• $$R\approx 8.315J\cdot \text{ K}^{ -1 }\cdot \text{ mol}^{ -1 }$$, ideal gas constant.
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