Why Is Average Velocity Calculated Differently in Various Scenarios?

[supernova88]

I'm trying to understand the kinematic equations well enough to explain to someone else, but the analogies I've come up with don't seem to be working well.

1) If a car travels 25 km/h for a certain distance, turns around in the opposite direction, and then turns back on its original path yet again before reaching its destination, then on a distance vs time graph its average velocity is found using the pythagorean theorem for how far the destination is from the origin versus the total time of the trip. This will be far less than 25 km/h. However, according to the equation (Vfinal + Vinitial)/2 = Vaverage, if its initial and final velocities are both 25 km/h in the same direction then the average velocity should be (25 + 25)/2 = 25 km/h. What is the discrepancy between these two versions of the average velocity? I believe the issue is (Vfinal + Vinitial)/2 only applies for cases of constant acceleration, which makes me wonder why we would say there is an average velocity at all if the velocity is constantly changing.

2) A space shuttle reaches orbital velocity of 28,000 km/h in 8.5 minutes (0.1417 h). On a velocity vs time graph this should form a triangle as velocity grows from 0 to 28,000 km/h. The distance should therefore be the area under the curve where the base (time) is 0.1417 h and the height (velocity) is 28,000 km/h. The area, and distance, is therefore 1/2(0.1417 h) (28,000 km/h) = 1983.8 km. However the space shuttle orbits at a height of 360 km give or take. I think the confusion is in the fact the space shuttle doesn't fly straight up but at an angle, so it travels 2,000 km but only as high as 360 km, but I'm not positive on this fact.

 

[jtbell]

I believe the issue is (Vfinal + Vinitial)/2 only applies for cases of constant acceleration,

Yes, this is why it doesn't apply to your situation (1).

which makes me wonder why we would say there is an average velocity at all if the velocity is constantly changing.

Imagine measuring and recording the instantaneous velocity (e.g. by looking at a car's speedometer) at frequent regular time intervals, say once per second during a one-hour trip. At the end of the trip, calculate the average of these measurements. (Of course, it's simpler if the trip is back and forth along a straight line; otherwise you have to deal with vector components.) This average will be very nearly equal to the simple ratio (net displacement)/(total elapsed time) which is the usual definition of "average velocity". The shorter the time intervals between measurements, the better the agreement.

2) [...] I think the confusion is in the fact the space shuttle doesn't fly straight up but at an angle, so it travels 2,000 km but only as high as 360 km, but I'm not positive on this fact.

The shuttle's path is curved: it starts out traveling vertically upwards, then curves towards the horizontal, parallel to the Earth's surface. The 2000 km is the distance along this path. (assuming constant magnitude of acceleration of course)

 

[supernova88]

Thanks for all the help. I thought I understood where I was confused but I needed some confirmation.

 

[haruspex]

It is better to distinguish between velocity (a vector) and speed (a scalar). Strictly speaking, the average velocity over some time interval is the displacement (a vector) divided by the time: $\frac{\int \vec v.dt}{\int dt}$. In your first example, the return trip reduces the displacement. Once back at the start position, the average velocity for the whole trip is zero.

For the rocket, again, you really mean speed, which is distance traveled along the path divided by time: $\frac{\int |\vec v|.dt}{\int dt}$.

 

[PJC01]

I can understand your confusion about average velocity and the discrepancies between different methods of calculating it. Let me try to clarify some of these points for you.

Firstly, the equation (Vfinal + Vinitial)/2 = Vaverage is indeed only applicable for cases of constant acceleration. This means that the velocity of the object is changing at a constant rate over time. In the case of the car traveling at a constant speed of 25 km/h and then turning around, the acceleration is not constant and therefore this equation cannot be used to calculate the average velocity. In such cases, we need to use the pythagorean theorem to calculate the average velocity, as you have correctly pointed out.

Secondly, the reason we still use the term "average velocity" in cases where the velocity is constantly changing is because it gives us a general idea of the overall speed of the object over the entire journey. It may not be an accurate representation of the actual speed at any given moment, but it still provides us with a useful measure.

Moving on to your second example of the space shuttle, you are correct in thinking that the confusion lies in the fact that it does not fly straight up but at an angle. This means that the distance traveled by the shuttle is not simply the height of the orbit, but also the horizontal distance it covers. This is why the area under the velocity vs time graph is used to calculate the distance traveled, taking into account both the height and horizontal distance.

I hope this helps to clarify your understanding of average velocity and its calculations. As with any scientific concept, it is important to consider the specific conditions and variables involved in order to accurately apply the relevant equations and calculations.