# Average Velocity confusion

1. Aug 11, 2014

### 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.

2. Aug 11, 2014

### Staff: Mentor

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

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.

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)

3. Aug 11, 2014

### supernova88

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

4. Aug 12, 2014

### 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 travelled along the path divided by time: $\frac{\int |\vec v|.dt}{\int dt}$.