The figure below shows the relation between time and the distance from a fixed point of a cyclist who is moving in a straight line. Using this information, find the magnitude of the average velocity vector of the cyclist.
Now, we need to be really careful here. We’re given a graph that shows the relation between time and distance, but we want to find the magnitude of the average velocity vector. Velocity is the rate of change of displacement of an object. And distance and displacement are however so slightly different. Displacement is the distance from a fixed point. And so if we think about the graph here, the cyclist may have traveled 80 meters and then another 30 meters. But actually, his displacement from the origin or the fixed point is only 50 meters.
And so we know that speed is distance traveled divided by time. And so we can say that the average velocity of the cyclist will be displacement over time. Now, this does tell us to find the magnitude of the vector. But of course, the magnitude of the vector is just its length, and so it’s absolutely fine just to consider displacement divided by time. We said that the displacement from the fixed point was 50 meters. The time it took the cyclist to achieve this was 10 seconds. So the average velocity and hence the magnitude of the average velocity vector is 50 divided by 10 which is simply equal to five.
We’re working in meters for the displacement and seconds for the time. And so the magnitude of the average velocity vector of the cyclist, which is simply his average velocity, is five meters per second.