Calculating New Northward Speed of Air Parcel

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SUMMARY

The discussion focuses on calculating the new northward speed of an air parcel initially moving at 10 m/s at 45°S after two hours, considering the effects of the Coriolis force. The Coriolis force is defined by the equations du/dt=fv and dv/dt=-fu, where f is the Coriolis parameter and Ω is the Earth's angular velocity. The key takeaway is that while the Coriolis force alters the direction of the parcel's velocity, it does not change the scalar speed. The term "northward speed" refers to the magnitude of the velocity component in the northward direction.

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Homework Statement



A parcel of air initially moves northward at a speed of 10 m/s at 45°S. After two hours, determine the new northward speed.

Homework Equations



Energy of a parcel= u^2+v^2

The Coriolis force accelerates a parcel such that, du/dt=fv and dv/dt=-fu
where f=2Ωsin[itex]\varphi[/itex] [itex]\varphi[/itex] =latitude and Ω=angular velocity of earth


The Attempt at a Solution



My first question about this problem is in regards to what it is asking for. It says to find the new northward speed of the parcel, however the Coriolis force only changes direction, not speed. So I then considered that the northward component of speed would change and that must be what the question is asking for. But then I thought about how speed is a scalar, not a vector, so there would be no "northward component" of speed because there simply aren't speed vectors. Please correct me if I'm wrong.
 
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Welcome to PF,

So you're just hung up on terminology? The "northward speed" means the speed in the northward direction, which is the magnitude of the northward component of the velocity.
 

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