22. B (9.8m/s/s ~ 10 m/s/s)
23. C
24. C
25. C
26. B
27. B
28. D
29.
a. A constant velocity is depicted by dots which are spaced the same distance apart.
b. An object which is accelerating in the same direction as the velocity (both are rightward in this case) is speeding up. And so as you trace your eye from left to right across the diagram (rightward), the dots should be spaced further and further apart to indicate that the object is speeding up.
c. An object that moves rightward (i.e., rightward velocity) and accelerating leftwards must be slowing down. And so as you trace your eye from left to right across the diagram (rightward), the dots should be spaced closer and closer together to indicate that the object is slowing down.
d. The dots begin by being equally spaced (constant speed) and close together (slow); then the spacing between dots become gradually further and further apart to indicated the speeding up nature of the motion.
e. The object is moving rightwards so you will need to trace your eye from left to right across the diagram (rightward). As you do you will note that the dots become closer and closer to indicate that the object is slowing down. Then the dots are piled onto the same location to indicate that the object is at rest. Finally, the dots are spread further and further apart; the rate at which the distance between dots is very gradual - consistent with the statement that the rate of acceleration is less than the original rate of deceleration.
30. There is no difference
31.
The governing principle in this problem is that the slope of a position-time graph is equal to the velocity of the object.
a. An object at rest (v = 0 m/s) is represented by a line with 0 slope (horizontal line).
b. An object with a constant, positive velocity is represented by a straight line (constant slope) which slopes upward (positive slope).
c. An object moving in the - direction and speeding up is represented by a line which slopes downward (- slope) and increases its steepness (increasing slope).
d. An object moving in the + direction and slowing down is represented by a line which slopes upward (+ slope) and increases its steepness (increasing slope).
e. An object moving in the positive direction at constant speed would be represented by a straight diagonal line (constant speed) which slopes upward (+ velocity). So in this case, there will be two straight diagonal lines; the second line will slope more than the first line.
f. An object moving in the negative direction with negative acceleration is speeding up (since the a vector is in the same direction as the motion). So on a p-t graph, this object will be represented by a line which slopes downwards (- velocity) and increases its slope over time (speeding up).
g. An object moving in the negative direction with positive acceleration is slowing down (since the a vector is in the opposite direction as the motion). So on a p-t graph, this object will be represented by a line which slopes downwards (- velocity) and levels off or becomes more horizontal over time (slowing down).
32.
The governing principle in this problem is that the slope of a velocity-time graph is equal to the acceleration of the object. Furthermore, a negative velocity would be a line plotted in the negative region of the graph; a positive velocity would be a line plotted in the positive region of the graph.
a. An object at rest (v = 0 m/s) is represented by a line located on the time axis (where v = 0 m/s).
b. An object moving in the positive direction at constant speed will be represented on a v-t graph by a horizontal line (slope = a = 0 m/s/s) positioned in the + velocity region.
c. An object moving in the negative direction and speeding up will be represented on a v-t graph by a sloped line located in the - velocity region. Since such an object has a - acceleration, the line will slope downwards (- acceleration).
d. An object moving in the negative direction and slowing down will be represented on a v-t graph by a sloped line located in the - velocity region. Such an object has a positive acceleration (since it is slowing down, the a vector will be in the opposite direction of the motion). The + acceleration would be consistent with a line that slopes upwards.
e. An object moving with a + velocity and a + acceleration would be represented on a v-t graph by a sloped line located in the + velocity region. The + acceleration would be consistent with a line that slopes upwards.
f. An object moving with a + velocity and a - acceleration would be represented on a v-t graph by a sloped line located in the + velocity region. The - acceleration would be consistent with a line that slopes downwards.
g. An object moving with a constant + velocity would be represented on a v-t graph by a horizontal line (slope = a = 0 m/s/s) located in the + velocity region. The slowing down to rest portion of the graph would be represented by a line which slopes downwards (- acceleration) towards the time axis.
33.
In the top graph, the object moves in the + direction with an acceleration from fast to slow until it finally stops. Then the object remains at rest. This means there is a positive velocity and a negative acceleration; the final velocity of the object is 0 m/s. So on a v-t graph, the line needs to be in the +velocity region and the slope needs to be - (for a negative acceleration). The line should end on the v=0 m/s axis (corresponding to its final rest position).
In the bottom graph, the object is moving in the - direction with a constant speed. It then gradually slows down until it stops; it then remains at rest. So on a v-t graph, there should be a horizontal line (a=0 m/s/s) in the -velocity region of the graph. Then the line should slope upwards (for the + acceleration that is characteristic of objects moving in the - direction and slowing down). The line should end on the v=0 m/s axis (corresponding to its final rest position).
34.
In the top graph, the object moves in the + direction with an acceleration from slow to fast. Then the object maintains a constant speed by walking in the same direction (+ direction). On a p-t graph, this would correspond to a line with positive slope (for moving in the + direction) which eventually straightens out into a diagonal line with constant slope (constant speed) in the positive direction (positive velocity).
In the bottom graph, the object is moving in the + direction and slowing down from a high speed to a slow speed until it finally changes direction; the object then moves in the - direction and speeds up. On a p-t graph, this would correspond to a line with positive slope (for moving in the + direction) which gradually levels off to a horizontal (for slowing down); then the line begins to slope downwards (for moving in the - direction) and gradually becomes steeper and steeper (for speeding up).
35.
Use area calculations to determine the displacement of the object.
| | | |
| Area of the pink triangle: A = 0.5*b*h A = 0.5*(5 s)*(10 m/s) A = 25 m | Area of the green rectangle: A = b*h A = (10 s)*(10 m/s) A = 100 m | Area of the blue triangle plus purple rectangle: A = 0.5*b1*h1 + b2*h2 A = 0.5*(5 s)*(5 m/s) + (5 s)*(5 m/s) A = 12.5 m + 25 m A = 37.5 m |
36.
It is a velocity graph; so merely read the velocity values off the graph.
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| v = 10.0 m/s |
|
37.
Do slope calculations to determine the acceleration. Slope is rise/run.
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a = (10 m/s)/(5 s) a = 2.0 m/s/s | a = slope = rise/run a = (0 m/s)/(10 s) a = 0.0 m/s/s | a = slope = rise/run a = (-5 m/s)/(5 s) a = -1.0 m/s/s |
38. During the first 5 seconds, the auto is moving with a + velocity and a + acceleration, increasing its speed from slow to fast. From 5 s to 15 s, the auto is moving with a constant speed and a zero acceleration. From 15 s to 20 s, the auto is moving with a +velocity and a - acceleration, decreasing its speed from fast to slow.
39.
In the first five seconds, the object is speeding up so the spacing between dots will gradually increase. In the next ten seconds (5-15 s) the object is maintaining a constant velocity so the spacing between dots remains the same. In the last 5 seconds (15-20 s) the object is slowing down so the spacing between dots is gradually decreasing.
For a position-time graph, the velocity is determined from a slope calculation. So in each case, two points must be picked and slope calculation must be performed.
a. Pick the two points: (0 s, 20 m) and (5 s, 10 m)
slope = rise/run = (-10 m)/(5 s) = -2.0 m/s
b. Pick the two points: (5 s, 2 m) and (10 s, 6 m)
slope = rise/run = (4 m)/(5 s) = +0.80 m/s
c. From 0 to 20 s, the slope is a constant value. So determining the slope at 13 seconds is as simple as merely determining the slope of the line choosing any two points. So just pick the two points: (0 s, 10 m) and (20 s, 50 m)
slope = rise/run = (40 m)/(20 s) = +2.0 m/s
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