Question

having problems staying logged in or are you experiencing issues? please visit our troubleshooting section for solutions. consider the dot diagrams for objects a, b, and c. the arrow represents the direction of motion. match the motion of objects a, b, and c to one of the lines on the graph. numbers can be used only one time. a matches graph: -- b matches graph: -- c matches graph: --

Explanation:

Step1: Analyze Object A

Object A's dots are close together (arrow left, so moving backward, position decreasing over time). The lower graph (pos'n vs t decreasing) has steeper slopes for faster speed. A has many dots (small time intervals, fast speed? Wait, no: dot spacing. Wait, dot diagrams: closer dots mean slower speed (more dots in same distance, so time between dots is less, speed = distance/time, so less time → higher speed? Wait, no: if moving left, position is decreasing. Let's see: Object A: dots are close (many dots), so time between dots is small, so speed is (distance between dots)/time. If dots are close, distance between is small, but time between is also small. Wait, maybe better: for the lower graph (position decreasing over time), steeper slope means faster speed (since slope is velocity, negative here). Object A: arrow left (position decreasing), dots are close (so speed: if dots are close, time between dots is short, so speed is (distance between dots)/time. If dots are close, distance is small, time is small, but maybe speed is fast? Wait, no: let's count dots. Object A: 10 dots? Object B: 5 dots. Object C: 4 dots. Wait, the direction: A is left (position decreasing), B and C are right (position increasing). So A must match a lower graph (4,5,6). B and C match upper graphs (1,2,3).

For A (left, position decreasing): the lower graph lines: 4,5,6. The slope (velocity) magnitude: steeper slope means faster speed. Object A has many dots (so time between dots is small, so speed is (distance between dots)/time. If dots are close, distance between is small, but time between is small. Wait, maybe the number of dots: A has more dots (so more time intervals? No, the length of the dot diagram is same? Maybe the spacing between dots: A's dots are close (small distance between), B's dots are spaced (medium), C's dots are far (large distance between). So for B (right, position increasing): speed is (distance between dots)/time. B has dots with medium spacing, C has large spacing. So C is faster than B. Upper graphs: 1 is steepest (fastest), 2 medium, 3 least steep (slowest). So C (large spacing, fast) matches 1, B (medium) matches 2, A (left, fast? Wait A's dots are close, so if moving left, speed: if dots are close, distance between is small, so speed is small? Wait no, maybe I got it reversed. Wait, dot diagram: each dot is a position at a time interval. So more dots in the same length means more time intervals, so time between dots is smaller. So speed = distance between dots / time between dots. So if dots are close (small distance), time between is small, so speed could be same? No, let's take an example: suppose the length of the dot diagram is L. For A: number of dots n_A, so number of intervals n_A - 1. Distance per interval: L/(n_A - 1). Time per interval: Δt. So speed v_A = (L/(n_A - 1))/Δt. For B: n_B dots, v_B = (L/(n_B - 1))/Δt. Since n_A > n_B > n_C, (n_A - 1) > (n_B - 1) > (n_C - 1), so L/(n_A - 1) < L/(n_B - 1) < L/(n_C - 1). So v_A < v_B < v_C. Wait, that makes sense: more dots mean smaller distance per interval, so slower speed. So A is slowest (left, position decreasing), B is medium (right, increasing), C is fastest (right, increasing). Upper graphs (position increasing): 1 is steepest (fastest), 2 medium, 3 least steep (slowest). So C (fastest) matches 1, B (medium) matches 2, A (slowest left) matches which lower graph? Lower graphs: 4,5,6. Slope magnitude: 6 is steepest (fastest left), 5 medium, 4 least steep (slowest left). Since A is slowest left, it should match 4? Wait no, earlier we said v_A < v_B < v_C. So A's speed (magnitude) is smallest, so slope magnitude (for left, negative slope) is smallest, so line 4 (least steep downward) matches A. B is medium right, matches 2. C is fastest right, matches 1. Wait, but let's check again. Object A: arrow left (position decreasing), dots close (slow speed, so small slope magnitude, so line 4 (least steep downward). Object B: arrow right, dots medium spacing (medium speed, line 2 (medium steep upward)). Object C: arrow right, dots far apart (fast speed, line 1 (steepest upward)).

Step2: Match A, B, C

- A (left, slow) → graph 4? Wait no, the lower graphs: 4,5,6. Wait the lower graph's y-axis is position, x is time. So as time increases, position decreases. The slope of 4 is least steep (smallest negative slope, so slowest speed left), 5 medium, 6 steepest (fastest left). Since A has many dots (slow speed left), it should match 4? Wait no, earlier calculation: v_A < v_B < v_C. So A's speed (magnitude) is smallest, so slope magnitude (|v|) is smallest, so line 4 (smallest |slope|) for A. B: medium speed right, line 2 (medium |slope|). C: fastest speed right, line 1 (largest |slope|). Wait but the problem says numbers can be used only once. Upper graphs: 1,2,3 (increasing, positive slope). Lower: 4,5,6 (decreasing, negative slope). A is decreasing, so lower graph. B and C increasing, upper. So A: lower graph (4,5,6). B: upper (1,2,3), C: upper (1,2,3). Since C is fastest, C→1. B is medium, B→2. Then A→4? Wait no, maybe I messed up the speed. Let's count the number of dots: A has 10 dots (9 intervals), B has 5 dots (4 intervals), C has 4 dots (3 intervals). So distance per interval: same length, so A: length /9, B: length/4, C: length/3. So speed: C > B > A. So C (fastest) → upper graph 1 (steepest), B (medium) → upper graph 2, A (slowest, left) → lower graph 4 (least steep downward). Wait but the lower graphs: 4,5,6. 4 is least steep, 6 is steepest. So A (slowest left) → 4? No, wait, if A is moving left, its position is decreasing over time. The lower graph's lines: 4 has the least steep slope (smallest negative velocity), 6 has the steepest (largest negative velocity). Since A is slowest (smallest |v|), it should have the least steep slope, so 4. B is medium right, so upper graph 2. C is fastest right, upper graph 1. Wait but let's check the dot spacing: A's dots are close (small distance between), so speed is small (since speed = distance/time, time is same for each dot? Maybe each dot is at equal time intervals. So time between dots is Δt (same for all). Then speed = distance between dots / Δt. So A: small distance → small speed. B: medium distance → medium speed. C: large distance → large speed. So for right-moving (B, C), speed: C > B. So upper graphs (position vs t, increasing): 1 is steepest (largest slope, fastest), 2 medium, 3 least steep (slowest). So C→1, B→2. For left-moving (A), speed is small (small |slope|), so lower graph (position vs t, decreasing): 4 is least steep (smallest |slope|), 5 medium, 6 steepest. So A→4? Wait no, the lower graphs: 4,5,6. Let's see the lower graph's lines: 4 has the shallowest slope (closest to horizontal), 6 has the steepest. So A (slowest left) → 4, B→2, C→1. But wait, the problem says "numbers can be used only one time". So upper graphs: 1,2,3. Lower:4,5,6. So A must be 4,5, or 6. B and C 1,2,3. So C (fastest) →1, B (medium)→2, A (slowest left)→4? Or maybe A is 6? Wait no, if A is moving left, and has many dots (small distance between), so speed is small, so slope is small (close to horizontal), so 4. Yes. So:

A matches graph: 4? Wait no, maybe I got the direction wrong. Wait the arrow for A is left, so position is decreasing. The lower graph's y-axis is position, x is time. So as time increases, position decreases. So line 4: position decreases slowly (small slope), line 6: position decreases fast (steep slope). Since A has many dots (small distance between, so small speed), it should be line 4. B has medium dots (medium speed right), line 2. C has few dots (large distance between, fast speed right), line 1.

Wait but let's check the original problem again. The dot diagrams:

A: many dots, arrow left.

B: medium dots, arrow right.

C: few dots, arrow right.

Graphs:

Upper (pos'n up with t): 1 (steep), 2 (medium), 3 (shallow).

Lower (pos'n down with t): 4 (shallow), 5 (medium), 6 (steep).

So speed (magnitude):

C (few dots, large distance) → fast → upper 1.

B (medium dots, medium distance) → medium → upper 2.

A (many dots, small distance) → slow → lower 4 (shallow down).

Yes, that makes sense. So:

A matches graph: 4? Wait no, the lower graphs are 4,5,6. Wait the numbers: upper graphs are 1,2,3 (increasing pos'n), lower 4,5,6 (decreasing pos'n). So A is decreasing, so 4,5, or 6. B and C increasing, 1,2,3.

So C (fastest) →1, B (medium)→2, A (slowest)→4? Or maybe A is 6? Wait no, speed magnitude: A is slow, so slope magnitude is small, so 4 (small slope magnitude). So:

A: 4? No, wait, maybe I mixed up. Let's think of velocity:

For right-moving (B, C), velocity is positive. Slope of pos'n vs t is velocity. So steeper slope → higher velocity.

C has larger spacing → higher velocity → slope 1 (steepest).

B has medium spacing → slope 2 (medium).

A is left-moving → velocity negative. Slope is negative. Steeper negative slope (more negative) → higher speed (magnitude). A has small spacing → small speed (magnitude) → slope is small negative (close to zero) → line 4 (shallowest negative slope).

Yes, so:

A matches graph: 4? Wait no, the lower graphs: 4,5,6. 4 is the shallowest (least steep) downward, 6 is steepest. So A (slowest left) →4, B (medium right)→2, C (fastest right)→1.

But wait, the problem says "numbers can be used only one time". So 1,2,3,4,5,6: each used once. So A:4, B:2, C:1? Or maybe A:6, B:3, C:1? No, B is medium, so 2. C is fast, 1. A is slow left, 4.

Alternatively, maybe I got the speed wrong. Let's count the number of dots:

A: 10 dots (9 intervals)

B: 5 dots (4 intervals)

C: 4 dots (3 intervals)

So time between dots: same Δt. So distance per interval:

A: d_A = L / 9

B: d_B = L / 4

C: d_C = L / 3

So speed:

v_A = d_A / Δt = L/(9Δt)

v_B = L/(4Δt)

v_C = L/(3Δt)

So v_C > v_B > v_A.

So for right-moving (B, C), velocity is positive. So slope (v) is positive. So steeper slope (higher v) → C (v_C) → slope 1 (steepest), B (v_B) → slope 2 (medium), A (v_A, negative) → slope 4 (shallowest negative, since |v_A| is smallest, so slope is closest to zero).

Yes, so:

A matches graph: 4? Wait no, the lower graphs are 4,5,6. 4 is the shallowest downward, 6 is steepest. So A (small |v|) →4, B (medium v)→2, C (large v)→1.

But let's check the answer. Maybe the correct matches are:

A: 6? No, that would be fast left, but A has many dots (slow). Wait, maybe I had the direction of the arrow wrong. Wait the arrow for A is left, so position is decreasing. The lower graph's lines: 4,5,6. If A is moving left quickly, it would have steep slope, but A has many dots (slow). So no, A is slow left, so shallow slope (4). B is medium right, medium slope (2). C is fast right, steep slope (1).

So:

A matches graph: 4

B matches graph: 2

C matches graph: 1

Wait but the problem says "numbers can be used only one time". So 1,2,3,4,5,6: A uses 4, B uses 2, C uses 1. Then 3,5,6 are left? No, wait, maybe I made a mistake. Wait the upper graphs are 1,2,3 (pos'n up), lower 4,5,6 (pos'n down). So A is down, so 4,5,6. B and C up, so 1,2,3. So C (fast)→1, B (medium)→2, A (slow)→4. Then 3,5,6 are unused? No, the problem says "match the motion of Objects A, B, and C to one of the lines on the graph. Numbers can be used only one time." So three objects, three graphs. Wait, there are six graphs? No, wait the image: upper graph has 1,2,3; lower has 4,5,6. So three objects, three graphs. So A must be one of 4,5,6; B and C one of 1,2,3. So three graphs: one from lower, two from upper. So:

C (fast right) →1 (upper steepest)

B (medium right) →2 (upper medium)

A (slow left) →4 (lower shallowest)

Yes, that uses 1,2,4. Wait but the problem says "numbers can be used only one time" – maybe each number is used once, so three numbers, one for each object. So A:4, B:2, C:1.

Answer:

A matches graph: 4
B matches graph: 2
C matches graph: 1