Does dx/dt Equal dv or Average Velocity?

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The discussion clarifies that dx/dt represents instantaneous velocity, defined mathematically as v(t) = dx(t)/dt. This is computed using the limit definition of the derivative, specifically v(t₀) = lim(t→t₀) (x(t) - x(t₀)) / (t - t₀). It is established that while dx/dt and dv can be equivalent under constant velocity conditions, they differ when velocity changes, with dx/dt indicating instantaneous velocity and v representing average velocity over an interval.

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quasi426
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Does dx/dt = dv or just v(avg.)?


Thanks
 
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There's no average,it's the function itself.

v(t)=:\frac{dx(t)}{dt}

We call it instant velocity,as we can use the definition of the derivative.We compute the "x" comp.of the velocity at the moment of time t_{0} by

v(t_{0})=:\lim_{t\rightarrow t_{0}} \frac{x(t)-x(t_{0})}{t-t_{0}}

or simply by plugging the time value in the velocity function itself.

Daniel.
 
for your question! The answer to this question depends on the context and the variables involved. In general, dx/dt represents the rate of change of position with respect to time, while dv represents the rate of change of velocity with respect to time. If we are dealing with a constant velocity, then dx/dt and dv would be the same, as the velocity is not changing. However, if the velocity is changing, then dx/dt would represent the instantaneous velocity at a particular point in time, while v would represent the average velocity over a given interval. So, it's important to specify which variable is being used in the equation and in what context.
 

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