What is the net displacement of the particle between 0 seconds and 80 seconds?
Velocity (m/s)
chhi NSON WAO
20
40
80
280
100
Кузорил
Time (s)
OA.
160 meters
OB.
80 meters
OC.
40 meters
OD.
20 meters
OE.
0 meters

Answer: 80

Explanation:

The question is incomplete, There is no specific time for the specific velocity, which is not provided in this question instead of giving the total time interval from 0sec to 80 sec. According to my assumption, I think the net displacement will be 20 m. The correct answer will be option D.  

Net displacement is the total change in the position of an object, taking into account both the magnitude and direction of its motion. It is a vector quantity and is expressed in terms of distance and direction.

In other words, it is the final position of an object relative to its initial position, irrespective of the path taken. It is calculated by finding the vector difference between the final and initial positions of the object.

For example, if an object moves 10 meters to the east and then 5 meters to the north of its starting position, then its net displacement would be the vector sum of these two displacements, which is the diagonal distance between its starting and final positions.

Here in the Question,

To calculate the net displacement of the particle between 0 seconds and 80 seconds based on the given options, we can use the fact that the net displacement is the same as the final position minus the initial position.

Let's assume that the initial position is zero.

Option A. If the net displacement is 160 meters, the final position would be 160 meters. However, looking at the velocities given, we can see that the particle goes back and forth, so it does not make sense for the net displacement to be larger than the maximum displacement in one direction.

Option B. If the net displacement is 80 meters, the final position would be 80 meters. This option could be possible if the particle moves in one direction for a while and then moves back to its initial position. This is incorrect.

Option C. If the net displacement is 40 meters, the final position would be 40 meters. This option is also possible if the particle moves in one direction for a shorter period of time. This is not relevant to the question.

Option D. If the net displacement is 20 meters, the final position would be 20 meters. This option could be possible if the particle moves in one direction for a very short period of time. This is most relevant to the Question.

Option E. If the net displacement is 0 meters, the final position would be the same as the initial position, which is zero. This option could be possible if the particle moves back and forth such that the positive displacements cancel out the negative displacements.

Therefore, based on the given options, the correct answer for the net displacement of the particle between 0 seconds and 80 seconds is option D: 20 meters.

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Relate acceleration qualitatively to a change in speed and/or direction as a result of a net force

A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward point A. What are
the total distance and total displacement of the boat?
OA. The total distance is 105 kilometers and the total displacement is 45 kilometers east.
OB. The total distance is 60 kilometers and the total displacement is 60 kilometers east.
OC. The total distance is 105 kilometers and the total displacement is 15 kilometers east.
OD. The total distance is 60 kilometers and the total displacement is 45 kilometers east.