Confused about speed and velocity...?

The magnitude of velocity is not always speed = distance/time traveled.

Here's an example. We have a circular track. The distance half way around the circumference from the start to end points is C/2 = pi R; where R is the radius of the track. We can run that distance in dT time. So our speed is distance/time = pi R/dT = speed to go half way around the circular track.

But the start and end points are 2 R, the diameter, apart. For example, if the start is at the 3 o'clock position the end point is at the 9 o'clock spot...half way around the clock (track). So the magnitude of the displacement is D = 2R and the direction is directly across the center of the circle from the start (3 o'clock) to the end point (9 o'clock).

So the velocity magnitude is displacement/time = 2 R/dT < pi R/dT = speed. dT is the same one as for the speed (it was the same run). And there you are. When the dT is the same, and the end and start points are the same, generally speaking the magnitude <= speed. For our example it was less than speed.

I disagree with Taryn’s answer. When you run around a circle at a speed of 5 m/s, your velocity is not 0 m/s. Velocity has magnitude and direction. When you run around a circle at a speed of 5 m/s, the magnitude of the velocity is 5 m/s and the direction of the velocity is always tangent to the circle. As you run a circle at a constant speed, the magnitude of the velocity is constant, but the direction is always changing.

Let me give you an example of problem where the displacement is 0, but the velocity is not 0. When a ball is thrown upward, its velocity decreases from the +49 m/s to 0 m/s. When the ball falls to its original height, its velocity increases from 0 to -49 m/s. The displacement is 0, but the velocity is constantly changing from +49 m/s to -49 m/s.

Your example below is similar to my example above.

Let’s assume that you run east at 5 m/s for 2 seconds, then turn around and run west at 5 m/s for 2 seconds. You ran 10 meters east and 10 meters west. So the total distance is 20 meters. Since your final position is the same as your initial position, your displacement is 0 meter. But, the only time that your velocity is 0 m/s is as you stop and turn around.

I still don't understand how the statement "the magnitude of the velocity vector is speed" can be accurate. I understand it is possible just that it doesn't seem very accurate...

Magnitude is the size of the velocity. When a car is moving east at 60 mph, the car’s speed is 60 mph. The size of the velocity is the same as the size of the speed. I hope this helps you understand why the magnitude of the velocity is the speed. And why there is no accuracy problem.

Magnitude is the number without the direction. If your velocity were -58km/h, then the "-" would be the direction, and the "58" would be the magnitude.

Basically speed is how fast something is going without worrying about the direction, and velocity is how fast something is going and which direction it's moving.

Now, the magnitude of velocity is not always speed, and this might be where you're confused.

If you ran a lap around a track, then your displacement would be 0 because you ended up where you started. This would make your velocity 0. However, you definitely were moving, and the rate at which you were doing this is your speed.

So the magnitude of velocity is only speed when you're moving in a straight line.

Speed and velocity can both be the same thing, if the speed is in 1 direction only, e.g. a long straight road.

But speed could be 60km/h. if a car drove 1 lap around a circular track in 1 hr., whereas velocity would be 0. The car is back where it started.

So similarly in the first part of your question, the average speed to a point distant from a start position could be 60kph., along a winding road.

But the straight line distance for that same car to the same destination might be 30km., so the average velocity would be 30kph., in that straight line.

Displacement is a straight line distance.