[ Questions #19  Questions #1045  Questions #4655  Questions #5672 ] 
10. If two displacement vectors of 6 meters and 8 meters (with varying directions) are added together, then the resultant could range anywhere between ___ meters and ___ meters.
a. 0, 48 
b. 1.33, 48 
c. 0, 14 
d. 2, 14 
e. ... nonsense! No such prediction can be made. 

f. ... nonsense! A prediction can be made but none of these choices are correct. 
Answer: D The vector sum of 6 meters and 8 meters will be the greatest if they are added together in the same direction; that would produce a resultant of 14 meters. The vector sum of 6 meters and 8 meters will be the smallest if they are added together in the opposite direction; that would produce a resultant of 2 meters. An animation depicting the various resultants of 6 meters added to 8 meters at varying directions is shown on a separate page. See animation. 
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11. Three vectors are added following the rules of vector addition. A fourth vector is drawn from the tail of the first vector to the head of the last vector. This fourth vector is referred to as ____.
a. the equilibrant b. the hypotenuse c. the resultant d. a mistake
Answer: C The resultant represents the result of adding two or more vectors. On a scaled vector addition diagram, the resultant is always drawn from the tail of the first vector to the head of the last vector. 
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The order in which vectors is added will effect the end result.
a. True b. False
Answer: B The order in which vectors is added does not effect the magnitude or direction of the resultant. A + B + C gives the same resultant as B + C + A and the same resultant as A + C + B. An animation depicting the the headtotail addition of five vectors in three different orders is shown on a separate page. See animation. 
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13. Vector A is directed northward and vector B is directed eastward. Which of the following vector addition diagrams best represent the addition of vectors A and B and the subsequent resultant?
Answer: E If the headtotail method (sometimes referred to as the triangle method) is being used, then the tail of B should be drawn starting at the head of A. Both diagrams D and E show this. Then the resultant should be drawn from the tail of A to the head of B (which is not shown in diagram D). There is also a parallelogram method for adding vectors. In this method, the tails of the two vectors are placed together (as in diagrams A and B). Then a parallelogram should be drawn with the two vectors forming the adjacent sides of the parallelogram. The resultant is drawn from the tails of the vectors to the opposite corner of the parallelogram. This is not done correctly on diagrams A and B. Diagrams C and F do not resemble any (accurate) vector addition method known to humankind. 
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14. When adding vector B to vector A geometrically (or graphically) using the head to tail method, the resultant is drawn from ____ to the ____.
a. head of A, tail of B b. tail of A, head of B c. head of B, tail of A d. tail of B, head of A
Answer: B Adding vector B to vector A is equivalent to saying A + B. In such an instance, A is drawn first, then B is drawn with its tail at the head of A. Finally, the resultant is drawn from the tail of the first vector (A) to the head of the last vector (B). 
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Use the following vector addition diagrams for questions #15#20.
15. Which one of the following vector addition equations is shown in Diagram 1?
a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these
16. Which one of the following vector addition equations is shown in Diagram 2?
a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these
17. Which one of the following vector addition equations is shown in Diagram 3?
a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these
18. Which one of the following vector addition equations is shown in Diagram 4?
a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these
Answers to questions #15#18: 15. B 16. A 17. E 18. G Vectors are added by a headtotail method and the resultant is drawn from the tail of the first vector to the head of the last vector. So if two vectors are added  say B is added to A (as in A + B)  then first A is drawn and the tail of B is placed at the head of A. The resultant is drawn with its tail at the tail of A and its head at the head of B. Thus, when the tails of two vectors are seen connected, one of the vectors is the resultant and the other is the first vector being added. And when the heads of two vectors are seen connected, one of the vectors is the resultant and the other is the second vector being added. These principles can be applied to answer the above questions. Note that in question 18, there is not a single vector which is drawn from the tail of one vector to the head of another vector. Thus, no resultant is drawn. One could say that the diagram shows that A + B + C = 0. 
19. Consider the magnitude and direction of vectors A and B as shown in Diagram 1 above. Which one of the following diagrams would represent B  A = R?
Answer: B The subtraction of A from B is equivalent to adding the negative of A to B. That is, B  A = B + (A). The negative of a vector is simply the same vector drawn in the opposite direction. The correct answer should be a diagram which shows B being drawn first. Then at the head of B, the tail of a vector pointing in the opposite direction of A should be drawn. This is shown in both diagrams B and C. The resultant should then be drawn from the tail of B to the head of A. This is not shown in diagram C. 
20. Consider the magnitude and direction of vectors B and C as shown in Diagram 2 above. Which one of the following diagrams would represent C  B = R?
Answer: C The subtraction of B from C is equivalent to adding the negative of B to C. That is, C  B = C + (B). The negative of a vector is simply the same vector drawn in the opposite direction. The correct answer should be a diagram which shows C being drawn first. Then at the head of C, the tail of a vector pointing in the opposite direction of B should be drawn. This is shown in both diagrams B and C. The resultant should then be drawn from the tail of C to the head of B. This is not shown in diagram B. 
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21. The vector sum (magnitude only) of 25.0 m, north + 18.0 m, East is ___ m.
a. 7.00 b. 21.5 c. 30.8 d. 35.8 e. 43.0 f. 54.2 g. 949 h. None of these
Answer: C These two vectors are directed at right angles to each other. In such instances, the vector sum can be determined using the Pythagorean theorem. The resultant (R) is equal to the square root of the sum of the squares of the two vectors being added. That is, R = Sqrt (A^{2} + B^{2}) where A and B are the two vectors being added together. Thus, R = 30.805 m = ~30.8 m 
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22. The vector sum (magnitude only) of 32.0 m, north + 41.0 m, west is ___ m.
a. 9.00 b. 36.5 c. 38.0 d. 52.0 e. 73.0 f. 128 g. 2705 h. None of these
Answer: D Like question #21 above, these two vectors are directed at right angles to each other. The Pythagorean theorem can be used to determine the resultant of their sum. The resultant (R) is equal to the square root of the sum of the squares of the two vectors being added. R = Sqrt [ (32.0 m)^{2} + (41.0 m)^{2} ] = Sqrt [ (1024 m^{2}) + (1681 m^{2}) ] = Sqrt (2705 m^{2}) R = 52.010 m = ~52.0 m 
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Use the diagram below to answer questions #23#28. Each square on the diagram represents a 20meter x 20meter area.
23. If a person walks from D to H to G to C, then the distance walked is ____ meters.
a. 128 b. 180 c. 401 d. 460 e. 480 f. 533 g. 620 h. None of these
Answer: F Distance is a scalar quantity and adding together three scalars is as simple as adding the three numbers arithmetically. However, since the first of the three legs of this trip is not due East, West, North or South, determining the distance for this leg requires that the Pythagorean theorem is used. That is, walking from D to H is equivalent to walking 140 meters east (7 squares) and 160 meters (8 squares) south. The distance from D to H is thus 212.6 meters  Sqrt [ (140 m)^{2} + (160 m)^{2} ]. The walk from H to G is a distance of 80 meters (4 squares) and the walk from G to C is 240 meters (12 squares). The total distance walked is the sum of 212.6m + 80 m + 240 m = 532.6 m = ~533 m. 
24. If a person walks from D to H to G to C, then the magnitude of the displacement is ____ meters.
a. 128 b. 180 c. 401 d. 460 e. 480 f. 533 g. 620 h. None of these
Answer: A Displacement is a vector quantity which indicates how far our of place an object is at the end of the motion relative to the beginning of the motion. Displacement does not depend upon the path taken from the beginning position to the final position, but only upon the distance of a vector drawn from start to finish. The resultant for the three segments of this walk is simply drawn from D to C. The distance of this resultant is found by using the Pythagorean theorem. The resultant stretches south 80 meters (4 squares) and west 100 meters (5 squares). The resultant has a magnitude of Sqrt [ (80 m)^{2} + (100 m)^{2} ] or 128 meters. 
25. If a person walks from D to H to G to C, then the direction of the displacement is ___ degrees (as measured counterclockwise from East).
a. 38.7 b. 51.3 c. 53.1 d. 216.9 e. 218.7 f. 231.3 g. 233.1 h. None of these
Answer: E As discussed in the previous problem, the resultant stretches south 80 meters (4 squares) and west 100 meters (5 squares). This is shown in the diagram at the right. These two parts of the resultant vector can be combined with a trigonometric function to determine the angle theta. The angle theta is approximately 38.7 degrees. This is not the direction of the resultant displacement but simply the angle between the displacement vector and due West. The convention used to express the direction of the vector is to measure the counterclockwise angle of rotation from due East. So in this instance, the direction is 180 degrees + 38.7 degrees or 218.7 degrees. 
26. If a person walks from H to E to C to G, then the distance walked is ____ meters.
a. 80.0 b. 240. c. 253 d. 333 e. 493 f. 560. g. 640. h. None of these
Answer: E Distance is a scalar quantity and adding together three scalars is as simple as adding the individual distances of the three legs is as simple as adding the distances. arithmetically. The problem is simplified in that the first two legs of the trip are along the same line segment. The stop at E can be disregarded since its is merely a point on the line from location H to location C. That is HE + EC = HC. Since the HC line segment does not stretch not due East, West, North or South, determining the distance for this leg requires that the Pythagorean theorem is used. That is, walking from H to C is equivalent to walking 240 meters west (12 squares) and 80 meters (4 squares) north. The distance from D to H is thus 253.0 meters  Sqrt [ (240 m)^{2} + (80 m)^{2} ]. The walk from H to E to C is a distance of 253 meters and the walk from C to G is 240 meters (12 squares). The total distance walked is the sum of 253 m + 240 m = ~493 m. 
27. If a person walks from H to E to C to G, then the magnitude of the displacement is ____ meters.
a. 80.0 b. 240. c. 253 d. 333 e. 493 f. 560. g. 640. h. None of these
Answer: A Displacement is a vector quantity which indicates how far our of place an object is at the end of the motion relative to the beginning of the motion. Displacement does not depend upon the path taken from the beginning position to the final position, but only upon the distance of a vector drawn from start to finish. The resultant for the three segments of this walk is simply drawn from H to G. The distance of this resultant is simply 80 meters (4 squares), north. 
28. If a person walks from H to E to C to G, then the direction of the displacement is ___ degrees (as measured counterclockwise from East).
a. 0 b. 18.4 c. 71.6 d. 90.0 e. 108.4 f. 161.6 g. 341.6 h. None of these
Answer: D The final destination G is due North of the initial location. Thus, the displacement is directed north (at 90 degrees). 
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Use the following diagram for questions #29#33. In the diagram, a riverboat is shown starting at position A on the east bank of a river. The boat heads towards position B (a point directly across the river from A) with a speed of 3.8 m/s. But because of a current with a speed of 1.8 m/s, the boat lands on the west bank of the river at position C, a location downstream from B. The width of the river (d_{across}) is 86.4 meters.
29. The magnitude of the resultant velocity of the boat is ____ m/s.
a. 1.8 b. 2.0 c. 3.35 d. 3.80 e. 4.20 f. 5.60 g.11.2 h. None of these
Answer: E This is a case of a boat crossing a river by use of its motor. The motor allows the boat to travel 3.8 meters towards the opposite shore in every second. The river is flowing south, carrying the boat a distance of 1.8 meters down the river every second. The resultant velocity is simply the vector sum of these two individual velocities. Since these two velocities are at right angles to each other, the vector sum can be determined using the Pythagorean theorem. v_{resultant} = Sqrt[ (3.8 m/s)^{2} + (1.8 m/s)^{2} ] v_{resultant} = 4.20 m/s 
30. The direction of the resultant velocity of the boat is ____ m/s.
a. 0 b. 18.4 c. 71.6 d. 90.0 e. 108.4 f. 161.6 g. 341.6 h. None of these
Answer: H The direction of the resultant velocity is in a southeastern direction. That puts the vector in the third quadrant with a direction somewhere between 180 degrees and 270 degrees. The exact angle can be determined if the angle theta is found using trigonometry. Theta is the angle that the resultant velocity (the red vector in the diagram at the right) makes with due West. This angle can be found suing the tangent function. The work is shown at the right. The angle theta is found to be 25.3 degrees. The actual direction as measured as the counterclockwise angle of rotation with due East is 180 degrees plus 25.3 degrees. This would be 205.3 degrees. 
31. The time required for the boat to cross the 86.4 m wide river is ___ seconds.
a. 4.20 b. 15.4 c. 20.5 d. 22.7 e. 48.0 f. None of these
Answer: D The motor allows the boat to travel 3.8 meters towards the opposite shore in every second. The boat has 86.4 meters to travel from shore to shore. (The presence of the current does not alter the width or shoretoshore distance.) The time to cross the river can be calculated from the river width and the boat velocity using the v = d/t equation. Rearranging the equation to solve for t yields t = (86.4 m) / (3.8 m/s) = 22.7 s 
32. Location C is the location where the boat ultimately lands on the opposite shore. What is the distance from location B to location C.
a. 37.0 b. 40.9 c. 78.1 d. 86.4 e. 95.6 f. 182 g. 202 h. None of these
Answer: B The distance that location C is downstream from B is mathematically related to the speed of the river and the time to cross the river. The distance can be calculated by multiplying the current speed by the time to cross the river. d_{downstream} = (1.8 m/s) • (22.7 s) d_{downstream} = 40.9 m 
33. If the current on a particular day was flowing with two times the velocity, then the time to cross the river would be ____.
a. two times greater b. onehalf as great c. greater, but not two times greater d. less, but not onehalf the original time e. the same as it was when the current flowed at 1.8
m/s.
Answer: E For this situation of a boat heading straight across the river, the current velocity is directed perpendicular to the boat velocity. These two components of the boats resulting motion are independent of each other. The boat velocity makes the sole contribution to the ability of the boat to cross the river. The river velocity only carries the boat southward down the river. So an alteration of the river velocity would have no effect on the time required for the boat to cross the river. 
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34. An object is undergoing free fall motion. As it falls, the object's ____.
a. speed increases 
b. acceleration increases 
c. both of these 
d. none of these 
Answer: A As an object freefalls, its velocity (and also its speed) changes by approximately 10 m/s every second. This means that the acceleration is a constant value of 10 m/s/s. An object has a changing speed (or velocity) and a constant acceleration if the speed changes by the same amount (a "constant amount") in each consecutive second of its motion. 
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35. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its velocity is _______.
a. entirely vertical 
b. entirely horizontal 
c. both vertical and horizontal 
d. not enough information given to know. 
Answer: B As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. At its peak, its vertical velocity becomes 0 m/s. At this instant in time, the velocity is entirely horizontal; there is no vertical component to the velocity. 
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36. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its acceleration is _______. (Neglect the effects of air resistance.)
a. entirely vertical 
b. entirely horizontal 
c. both vertical and horizontal 
d. not enough information given to know. 
Answer: A As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. At the peak and everywhere throughout the trajectory, there is a vertical (downward) acceleration. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force  downward. 
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37. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, the net force acting upon it is _______. (Neglect the effects of air resistance.)
a. entirely vertical 
b. entirely horizontal 
c. both vertical and horizontal 
d. not enough information given to know. 
Answer: A A projectile is an object upon which the only force is gravity. Since no other forces act upon the object, the net force would be downward. 
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38. At what point in its path is the horizontal component of the velocity (v_{x}) of a projectile the smallest?
a. The instant it is thrown. 
b. Halfway to the top. 
c. At the top. 
d. As it nears the top. 
e. It is the same throughout the path. 
Answer: E As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. Having a constant horizontal velocity, there is no point along the trajectory where the v_{x} value is smaller than at other points. 
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39. At what point in its path is the vertical component of the velocity (v_{y}) of a projectile the smallest?
a. The instant it is thrown. 
b. Halfway to the top. 
c. At the top. 
d. As it nears the top. 
e. It is the same throughout the path. 
Answer: C As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. During the upward portion of its trajectory, the v_{y} continuously decreases until it becomes 0 m/s at the peak. Thus, the v_{y} is as small as it will ever be when it is at the peak of the trajectory. 
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40. An airplane that flies at 100 km/h in a 100 km/h hurricane crosswind has a speed (relative to the ground) of ____.
a. 0 km/h 
b. 100 km/h 
c. 141 km/h 
d. 200 km/h 
Answer: C When an object such as a plane or a boat moves within a medium which is moving relative to the ground, the speed of the plane or boat (as measured by its speedometer) will not be the same as the speed as measured by a person on the ground (ground speed). The ground speed (i.e., resultant velocity produced by the combination of the plane's speed and the wind speed) can be determined by adding the plane speed and wind speed as vectors. In this case, the two vectors are at right angles, so the resultant velocity can be determined using the Pythagorean theorem. R = SQRT( (100 km/hr)^{2} + (100 km/hr)^{2} ) R = 141 km/hr 
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41. An airplane moves at 141 km/h toward the northeast (45 degrees). What is its component velocity in the northward direction?
a. 41 km/h 
b. 100 km/h 
c. 110 km/h 
d. 141 km/h 
Answer: B To determine the component of a vector in a given direction, vector resolution principles must be applied. This velocity vector is directed at 45 degrees and has magnitude of 141 km/hr. The vertical component (northward) of this vector can be found using the sine function. v_{y} = (141 km/hr)*sine(45 degrees) v_{y} = 100 km/hr 
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42. Roll a bowling ball off the edge of a table. As it falls, its horizontal component of velocity ___.
a. decreases 
b. remains constant 
c. increases 
Answer: B Once the ball leaves the table's edge, it becomes a projectile. As it falls, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force  downward. 
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43. A bullet is fired horizontally and hits the ground in 0.5 seconds. If it had been fired with twice the speed in the same direction, it would have hit the ground in ____. (Assume no air resistance)
a. less than 0.5 s. 
b. more than 0.5 s. 
c. 0.5 s. 
Answer: C Once the bullet leaves the muzzle, it becomes a projectile (assuming no air resistance). As it falls, its horizontal velocity remains constant while its vertical velocity decreases. The force of gravity acts upon the bullet to cause its downward acceleration. The motion of the bullet in the downward direction is independent of the motion in the horizontal direction. That is to say, any alteration in a horizontal aspect of its motion will not effect the motion in the vertical direction. The time to fall vertically to the ground is not effected by the horizontal speed of the projectile. It would still take 0.5 seconds to fall to the ground from this height regardless of the horizontal speed. 
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44. A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. For the same speed, what other projection angle would produce the same downrange distance?
a. 30 degrees. 
b. 45 degrees. 
c. 50 degrees. 
d. 75 degrees 
e. 90 degrees. 
Answer: D For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Any two launch angles which are separated from 45 degrees by the same amount (for example, 40 degrees and 50 degrees, 30 degrees and 60 degrees and 15 degrees and 75 degrees) will provide the same horizontal displacement. 
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45. Two projectiles are fired at equal speeds but different angles. One is fired at angle of 30 degrees and the other at 60 degrees. The projectile to hit the ground first will be the one fired at (neglect air resistance) ____.
a. 30 degrees 
b. 60 degrees 
c. both hit at the same time 
Answer: A For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Launch angles greater than 45 degrees result in large vertical components of velocity; these stay in the air longer and rise to higher heights. Launch angles less than 45 degrees result in small vertical components of velocity; these do not rise as high and end up falling to the ground in shorter times. 
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Last update: 4/30/02