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KEY CONCEPT FIGURE 2.26 Determining the sign of the acceleration. The object is moving to the right, so y >0. Because it is speeding up, The object is moving to the left, so v <0. Because it is slowing down, its acceleration vector points opposite to its velocity (i.e., to the right), so a, > 0. The object is moving to the right, so v > 0. Because it is slowing The object is moving to the left, so v, < 0. Because it is speeding up, its acceleration vector points in the same direction as its velocity (i.e., to the left), so a, <0. its acceleration vector points in the same direction as its velocity (i.e., to the right), so a, > 0. down, its acceleration vector points opposite to its velocity (i.e., to the left), so az < 0. Positive ax, positive slope V Positive ax, Negative a,, negative slope Negative a,, negative slope positive slope Vx Vy Vy t А. С. D. B.

KEY CONCEPT FIGURE 2.26 Determining the sign of the acceleration. The object is moving to the right, so y >0. Because it is speeding up, The object is moving to the left, so v <0. Because it is slowing down, its acceleration vector points opposite to its velocity (i.e., to the right), so a, > 0. The object is moving to the right, so v > 0. Because it is slowing The object is moving to the left, so v, < 0. Because it is speeding up, its acceleration vector points in the same direction as its velocity (i.e., to the left), so a, <0. its acceleration vector points in the same direction as its velocity (i.e., to the right), so a, > 0. down, its acceleration vector points opposite to its velocity (i.e., to the left), so az < 0. Positive ax, positive slope V Positive ax, Negative a,, negative slope Negative a,, negative slope positive slope Vx Vy Vy t А. С. D. B.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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An elevator is moving downward. It is slowing down as it approaches the ground floor. Adapt the information in 2.26 to determine which of the following (STT 2.5) velocity graphs best represents the motion of the elevator.

KEY CONCEPT FIGURE 2.26 Determining the sign of the acceleration. The object is moving to the right, so y >0. Because it is speeding up, The object is moving to the left, so v <0. Because it is slowing down, its acceleration vector points opposite to its velocity (i.e., to the right), so a, > 0. The object is moving to the right, so v > 0. Because it is slowing The object is moving to the left, so v, < 0. Because it is speeding up, its acceleration vector points in the same direction as its velocity (i.e., to the left), so a, <0. its acceleration vector points in the same direction as its velocity (i.e., to the right), so a, > 0. down, its acceleration vector points opposite to its velocity (i.e., to the left), so az < 0. Positive ax, positive slope V Positive ax, Negative a,, negative slope Negative a,, negative slope positive slope Vx
Transcribed Image Text:KEY CONCEPT FIGURE 2.26 Determining the sign of the acceleration. The object is moving to the right, so y >0. Because it is speeding up, The object is moving to the left, so v <0. Because it is slowing down, its acceleration vector points opposite to its velocity (i.e., to the right), so a, > 0. The object is moving to the right, so v > 0. Because it is slowing The object is moving to the left, so v, < 0. Because it is speeding up, its acceleration vector points in the same direction as its velocity (i.e., to the left), so a, <0. its acceleration vector points in the same direction as its velocity (i.e., to the right), so a, > 0. down, its acceleration vector points opposite to its velocity (i.e., to the left), so az < 0. Positive ax, positive slope V Positive ax, Negative a,, negative slope Negative a,, negative slope positive slope Vx
Vy Vy t А. С. D. B.
Transcribed Image Text:Vy Vy t А. С. D. B.
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