Physics Questions

1)
The min. displacement=-5-3-4=-12cm
The max. displacement=5+3+4=12cm
Therefore, the range of the displacement：
-12cm＜displacemet＜12cm
So, (1) and (2) are correct.

(3) is incorrect.Since distance is not a vector, distance is the length you travel, in this case, the distance must be equal to 12cm.

2)
Average speed=Total distance/time used
Average speed=(80+60)/(2+3)=28m/s

Average velocity=Displacement/time used
Average velocity=RT/5
I think it should have a figure to show the length of RT, right?

I believe you confusing the concepts of displacement and distance, don't you?
What you have to know is that displacement just consider the starting point to the final point while the distance consider all the length the object travel.

For instance, I move forwards 10m and then move backwards 8m, what is the displacement I travelled?My displacement is 10-8=2m, I am at the position which is in front of my starting point 2m.However, the distance I travelled is 8+10=18m, I move 18m in total.

2007-09-20 18:51:43 補充：
Q1 statement 1 is correct.Just try your very best to think, in this case, no direction is told, the ant can travel the 5cm, 3cm, 4cm in all direction, then if the ant move forwards 5cm in the first interval.

2007-09-20 18:52:02 補充：
Then keep on moving 1cm and backwards 2cm in the second interval, now the displacement is 4cm(5cm 1cm-2cm), in the third interval the ant move backwards 4cm, then the total displacement is zero(4cm-4cm).

2007-09-20 18:54:39 補充：
The figure：backward＜－－－－－＞ forwardAnt＝＝＝＝＝＞5cm　in the first interval＝＞1cm，＜＝＝2cm in the second interval＜＝＝＝＝4cm in the third intervalThen the displacement=5 1-2-4=0cm