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## Homework Statement

there are four moving objects, whose graphs are pictured in orange color.

I forgot to mention that the image, from the book, shows

(y-axis) is denoted by x metres.

The (x axis) is denoted by t seconds.

One could say that these hint towards the idea, that distance travelled was the requested characteristic by the questioner.

Therefore the coordinate system is place, with respect to time (I think this is what the book is telling me)

a) find which object has moved the most during 10 seconds.

b) find which object has moved the least during 10 seconds.

## Homework Equations

v=s/t velocity

v=d/t speed

## The Attempt at a Solution

objects shall be from upper left-->upper right-->lower left-->lower right

a; b; c; d;

It is difficult to say how one should interpret the graph I think. I think I calculated the velocities wrong in the image.

One could say I suppose that one should use speed equation instead of the velocity equation.

Objects A, and B have been stagnant in their movement because the place is not increasing with time. However A and B distance begin from a number which is >0. 30m distance was already clocked-in at 0 seconds. In case of B, we can see that 3m distance was already clocked in at 0 seconds.

One could argue that within the time period of 0-10s, the already clocked-in amount of distance should be summed up and counted. If you have a foot race with another person for 30m. Let's say that runner R starts the race from 0m marker. However, runner S starts the race from 0+30m marker. Of course, Runner S is at a great advantage. Runner S wins the competition instantly. Then again the physical distance that Runner S runs, is actually only 0m. Seems kind of a silly idea to have a competition like that. But I'm little bit stumped about the question because of that.

A seems to be the graph with the most distance. One can see from the graph that 0-10s is the timeline.

We can see that 30m is the beginning of the distance. We can see that the object is not moving in the cases of A and B. And they are remaining still at the distances of 30m and 3m respectively.

However the graph starts at 30m already clocked-in as distance. One could argue either that the distance is 30m because it was already clocked-in. Or, one could argue that no additional movement happened. Because no additional distance happened.