E1T3 PreAP Physics Numbers 1-5 2016

1. Which of the following situations represents a positive displacement of a carton? Assume positive position is measured vertically upward along a y-axis.

A. A delivery person waiting for an elevator lowers a carton onto a dolly.

B. When the elevator doors open, the delivery person lifts the dolly over the threshold of the elevator.

C. The delivery person pushes the dolly to the back of the elevator while pressing a floor button.

D. The door closes and the elevator moves from the 10th to the 4th floors.

In this situation, the carton is being lifted vertically upward by the delivery person. This represents a positive displacement along the y-axis, as the position of the carton is increasing in the upward direction.

Explanation

In this situation, the carton is being lifted vertically upward by the delivery person. This represents a positive displacement along the y-axis, as the position of the carton is increasing in the upward direction.

2. What is the maximum negative displacement a dog could have if it started its motion at +3 m?

A. +7 m

B. +3 m

C. -3 m

D. -7 m

The maximum negative displacement a dog could have if it started its motion at +3 m is -3 m. Displacement refers to the change in position of an object, and negative displacement indicates that the object has moved in the opposite direction of its initial position. In this case, starting at +3 m, the dog can move up to 3 meters in the negative direction, resulting in a maximum negative displacement of -3 m.

Explanation

The maximum negative displacement a dog could have if it started its motion at +3 m is -3 m. Displacement refers to the change in position of an object, and negative displacement indicates that the object has moved in the opposite direction of its initial position. In this case, starting at +3 m, the dog can move up to 3 meters in the negative direction, resulting in a maximum negative displacement of -3 m.

3. What is the correct description of any change of position farther to the right of zero?

A. positive displacement

B. positive distance

C. positive position

D. positive change of displacement

A positive displacement refers to any change of position that is farther to the right of zero. This means that the object or point has moved in the positive direction along the number line. Displacement specifically refers to the change in position from an initial point to a final point, and in this case, the displacement is positive because it is to the right of zero.

Explanation

A positive displacement refers to any change of position that is farther to the right of zero. This means that the object or point has moved in the positive direction along the number line. Displacement specifically refers to the change in position from an initial point to a final point, and in this case, the displacement is positive because it is to the right of zero.

4. A dog walks from +4 m to +2 m. Which of the following statements is true about the dog's motion?

A. xi = +2 m

B. xf= +2 m

C. delta x = +2 m

D. vavg = 2 m/s

The statement "xf= +2 m" is true about the dog's motion. This means that the final position of the dog is +2 m.

Explanation

The statement "xf= +2 m" is true about the dog's motion. This means that the final position of the dog is +2 m.

5. Rank in from high to low the displacements of objects having the following pairs of average velocity and time.I.v_avg = +2.0 m/s, Dt = 2.0 sII. v_avg = +3.0 m/s, Dt = 2.0 sIII. v_avg = +3.0 m/s, Dt = 3.0 s

A. I, II, III

B. II, III, I

C. II, I, III

D. III, II, I

The displacement of an object is calculated by multiplying the average velocity by the time interval. In this case, the displacement for each pair can be calculated as follows:

I. displacement = (2.0 m/s) * (2.0 s) = 4.0 m
II. displacement = (3.0 m/s) * (2.0 s) = 6.0 m
III. displacement = (3.0 m/s) * (3.0 s) = 9.0 m

Therefore, the correct order from high to low displacement is III, II, I.

Explanation

The displacement of an object is calculated by multiplying the average velocity by the time interval. In this case, the displacement for each pair can be calculated as follows:

I. displacement = (2.0 m/s) * (2.0 s) = 4.0 m
II. displacement = (3.0 m/s) * (2.0 s) = 6.0 m
III. displacement = (3.0 m/s) * (3.0 s) = 9.0 m

Therefore, the correct order from high to low displacement is III, II, I.

E1t3 Preap Physics Numbers 1-5 2016