E1t7 PBL Regular Preap Physics 2016

The velocity is constant when the graph is a horizontal line, indicating that the object is moving at a constant speed. In the given options, Interval 2 and Interval 5 are the only intervals where the graph shows a horizontal line, suggesting that the velocity is constant during these intervals. Therefore, the correct answer is Interval 2 and Interval 5 only.

The jogger has a constant positive velocity during the interval BC because the jogger is moving in the positive direction without changing their speed.

The jogger is speeding up during the interval EF because the letters represent points along a path. The jogger is moving from point E to point F, indicating an increase in speed.

During the interval DE, the jogger is at rest.

The object is slowing down during Interval 3 and Interval 6 only because the graph shows a decrease in velocity during these intervals. This indicates that the object is experiencing deceleration or negative acceleration, which means it is slowing down.

The graph illustrates that the acceleration is not constant throughout the given time period. It shows that the acceleration changes in magnitude and direction, indicating that it varies.

The jogger is slowing down during the interval CD.

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The jogger has a constant negative velocity during the interval FG because the velocity is negative and remains constant throughout this interval. This means that the jogger is moving in the opposite direction of the positive axis and at a constant speed.

The object is traveling west during Interval 4, Interval 5, and Interval 6 only because in these intervals, the object's position on the graph is decreasing from left to right, indicating a westward direction of travel.

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The jogger travels south during the interval FG because the given options represent different intervals. Since FG is the only interval that indicates a southward direction, it can be inferred that the jogger travels south during that interval.

The graph shows that both the acceleration at A and the acceleration at B are zero. This means that there is no change in velocity at either point A or point B. The absence of acceleration indicates that the object is either at rest or moving with a constant velocity at both points.

The object is speeding up during Interval 1 and Interval 4 only because in these intervals, the velocity of the object is increasing. This can be observed from the graph where the slope of the line representing the object's velocity is positive during these intervals.

The speed of the jogger is decreasing during the intervals AB and CD. This means that the jogger is slowing down during these intervals.

The magnitude of the jogger's acceleration is the greatest during the interval EF. This can be inferred because EF is the steepest part of the graph, indicating a rapid change in velocity. The greater the change in velocity over a given time interval, the greater the acceleration. Therefore, the magnitude of the jogger's acceleration is the highest during the interval EF.