Time rate of change of air pressure

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SUMMARY

The time rate of change of air pressure is analyzed in the context of a ship sailing eastward at 10 km/hour, experiencing a pressure drop of 30 pascals over 2 hours. The air pressure decreases at a rate of 2.5 pascals per kilometer. Using the multivariate chain rule, the equation \(\frac{dp}{dt} = \frac{\partial p}{\partial t} + \frac{\partial p}{\partial x}\frac{dx}{dt}\) is applied to determine the time rate of change of air pressure at the island. The calculated result indicates that the time rate of change of air pressure on the island is -15 Pa/hr.

PREREQUISITES
  • Understanding of basic calculus, specifically the multivariate chain rule
  • Familiarity with units of pressure, particularly pascals
  • Knowledge of kinematics, particularly speed and distance relationships
  • Basic understanding of atmospheric pressure concepts
NEXT STEPS
  • Study the application of the multivariate chain rule in different contexts
  • Learn about atmospheric pressure variations and their implications
  • Explore the relationship between speed, distance, and pressure changes in fluid dynamics
  • Investigate the effects of altitude on air pressure and its rate of change
USEFUL FOR

Students in physics or engineering, meteorologists, and anyone interested in understanding the dynamics of air pressure changes in relation to movement and time.

BondKing
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Air pressure decreases at a rate of 2.5 pascals per kilometer in the eastward direction. In addition, the air pressure is increasing at a constant rate with respect to time everywhere. A ship sailing eastward at 10 km/hour past an island takes barometer readings and records a pressure drop of 30 pascals in 2 hours. Estimate the time rate of change of air pressure on the island. (A pascal is a unit of air pressure.)

What is time rate of change of air pressure?

Attempt:

Basically tried everything possible

10 x 2 = 20 km

30 Pa / 2 hrs = 15 Pa / hr

2.5 Pa x 20 km = 50 Pa
 
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This is an exercise in using the multivariate chain rule in the form \frac{dp}{dt} = \frac{\partial p}{\partial t} + \frac{\partial p}{\partial x}\frac{dx}{dt} where x(t) is the position of the ship. You are trying to find \frac{\partial p}{\partial t} which is stated to be constant.
 
-15 = ∂p/∂t + (-2.5)(10) ?
Is the unit for the answer Pa/hr?
 
BondKing said:
-15 = ∂p/∂t + (-2.5)(10) ?
Is the unit for the answer Pa/hr?
Yes
 

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