Why is Average Velocity Defined Differently in Calculus?

[vanmaiden]

Homework Statement

When I looked a physics for a while, I learned that average velocity is the sum of the initial and final velocities divided by two. However, I looked through one of my calculus books and it defined average velocity as the change in displacement over the change in time. Could someone please explain why this is?

Homework Equations

$\frac{V_{2}+V_{1}}{2}$

$\frac{\Delta D}{\Delta T}$

The Attempt at a Solution

I did some research and wasn't able to find much. I was just told what the calculus definition was and I already knew that.

 

[SammyS]

The average velocity is defined as the displacement divided by the elapsed time.

If acceleration is constant during the time interval under consideration, then the average velocity may be calculated by taking the average of the initial & final velocities.

 

[vanmaiden]

The average velocity is defined as the displacement divided by the elapsed time.

If acceleration is constant during the time interval under consideration, then the average velocity may be calculated by taking the average of the initial & final velocities.

Thank you. That makes a great deal of sense. Very impressive degrees by the way.

 

[Ray Vickson]

If the acceleration is not constant, the 50-50 average of the initial and final velocities is not always equal to the true average velocity (the latter being displacement/time *by definition*). Note also that you need to distinguish between velocity and speed. For example, if I drive from A to B in 1 hour and immediately turn around and drive back from B to A in 1 hour, my average velocity = 0, irrespective of details. However, computing my average speed over those 2 hours would require details about acceleration, decelleration and the like.

RGV

 

[vanmaiden]

For example, if I drive from A to B in 1 hour and immediately turn around and drive back from B to A in 1 hour, my average velocity = 0, irrespective of details. However, computing my average speed over those 2 hours would require details about acceleration, decelleration and the like.RGV

Would you have an average velocity of 0 because the return trip velocity is negative?

 

[SammyS]

It's zero because the displacement from A to A is zero.

 

[HallsofIvy]

Be careful about the distinction between "velocity" and "speed". In one dimensional problems "velocity" is a signed number (in two or three dimensions, it is a vector) while "speed" is non-negative- the magnitude of the velocity.

If the distance from A to B is 200 m and you travel from A to B in 4 seconds and you travel from B back to A in 3 seconds, your average velocity, from A to B is 50 m/s while your average velocity from B to A is -66 2/3 m/s. Your average velocity for the entire trip is 0 because you are right back where you started. Your average speed from A to B is 50 m/s while your average speed from B to A is +66 2/3 m/s. Your average speed for the entire trip is 400/7= 56 1/7 m/s because you did a total of 400 m in 7 seconds.