Question

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's Motion

Object A has dots evenly spaced, so it moves at a constant velocity. A constant velocity on a position - time graph is a straight line with a constant slope. Looking at the graphs, graph 2 (in the first set) or graph 6? Wait, no, let's re - examine. Wait, the first position - time graph (top right) has graph 2 as a straight line (constant slope), graph 1 is a curve (decreasing slope, slowing down), graph 3 is a curve (decreasing slope, speeding up? No, wait, position - time graph slope is velocity. For Object A, equal spacing means constant velocity, so a straight line. Let's check the second graph (bottom right). Graph 6 is a straight line with negative slope? No, the arrow for motion is to the right, so positive direction. Wait, maybe the top graph: graph 2 is a straight line (constant positive slope), so Object A (constant velocity) matches graph 2? Wait, no, let's check Object B: dots getting closer, so slowing down (decreasing velocity, so position - time graph has decreasing slope). Object C: dots getting farther, so speeding up (increasing velocity, position - time graph has increasing slope).

Wait, let's correct:

- Object A: Dots are equally spaced. So velocity is constant. On a position - time graph, constant velocity is a straight line (linear). So we need a linear graph. Looking at the graphs: top right has graph 2 (linear, positive slope), bottom right has graph 6 (linear, negative slope) but motion is to the right (positive direction), so graph 2 (top right) or maybe graph 6 is negative direction. Wait, the bottom right graph has time increasing to the left? Wait, the arrow on the bottom time axis is to the left, so time is decreasing? No, maybe it's a typo, but usually time increases to the right. Assuming the top graph has time increasing to the right (t axis), and bottom graph maybe time increasing to the left (but that's unusual). Let's go back.

Dot diagrams:

- Object A: Equal spacing → constant velocity (uniform motion). Position - time graph: straight line (linear), slope = velocity (constant).

- Object B: Dots getting closer (in the direction of motion) → velocity decreasing (decelerating). Position - time graph: curve with decreasing slope (since slope = velocity, velocity decreasing means slope decreasing).

- Object C: Dots getting farther (in the direction of motion) → velocity increasing (accelerating). Position - time graph: curve with increasing slope (slope = velocity, velocity increasing means slope increasing).

Now looking at the top graph (Pos'n vs t, t increasing to the right):

- Graph 1: curve, slope decreasing (decelerating? No, wait, position vs time: if slope is velocity, a curve with decreasing slope (as t increases, slope gets smaller) means velocity decreasing. Wait, no, for a position - time graph, the slope is $v=\frac{\Delta x}{\Delta t}$. If the curve is concave down (like graph 1), the slope is decreasing (velocity decreasing). If concave up (graph 2? No, graph 2 is straight line, graph 1 is concave down, graph 3 is concave up? Wait, the top graph:

- Graph 1: concave down (slope decreasing)

- Graph 2: straight line (slope constant)

- Graph 3: concave up (slope increasing)

Bottom graph (Pos'n vs t, t increasing to the left? Unusual, but maybe it's a different time axis. But let's assume the top graph is standard (t right).

So:

- Object A (constant velocity) → graph 2 (straight line, constant slope)

- Object B (decelerating, velocity decreasing) → graph 1 (concave down, slope decreasing)

- Object C (accelerating, velocity increasing) → graph 3 (concave up, slope increasing)

Wait, but the bottom graph has graphs 4,5,6. Graph 6 is straight line (negative slope), graph 4 is concave up, graph 5 is concave down. But maybe the first set (top) is for positive direction motion, and the second set (bottom) is for negative? No, the dot diagrams have motion to the right (arrow right), so positive direction.

So:

A matches graph 2? Wait, no, maybe I got the graphs wrong. Wait, the problem says "Numbers can be used only one time". The numbers are 1,2,3,4,5,6.

Wait, let's re - express:

Dot diagram analysis:

- Object A: Equal spacing → constant velocity → position - time graph is linear (straight line) → so graph 2 (top) or graph 6 (bottom). But graph 6 is straight line with negative slope (decreasing position), but Object A is moving right (increasing position), so graph 2 (top, positive slope) or maybe graph 6 is negative direction. Wait, maybe the bottom graph's time axis is reversed (arrow to the left means time is decreasing, so as time decreases, position changes). But this is confusing. Alternatively, maybe the first three graphs (1,2,3) are for one set, and 4,5,6 for another, but the problem says "one of the lines on the graph" (all 6 lines).

Wait, let's try again:

- Object A: constant velocity → linear graph (slope constant) → among 1 - 6, graph 2 (top) and graph 6 (bottom) are linear. Graph 2 has positive slope (increasing position), graph 6 has negative slope (decreasing position). Since Object A is moving right (increasing position), graph 2? No, maybe graph 6 is for negative direction, but Object A is moving right. Wait, maybe the bottom graph's time is increasing to the left, so as time (t) increases (moving left on the axis), position changes. So for graph 6 (bottom), it's a straight line with negative slope (as t increases left, position decreases), which would be moving left, but Object A is moving right. So maybe graph 2 (top) is correct for A.

- Object B: dots getting closer (slowing down, velocity decreasing) → position - time graph has decreasing slope (concave down). Graph 1 (top) is concave down, graph 5 (bottom) is concave down. Let's see, graph 1 (top) has position starting high, decreasing? No, graph 1 (top) starts at high position, then the slope decreases (velocity decreases), but if moving right, position should increase. Wait, maybe I have the position axis wrong. Maybe "Pos'n" is position, and for moving right, position increases with time. So:

- Object A: position increases at constant rate → graph 2 (top, straight line, positive slope)

- Object B: position increases at decreasing rate (slowing down) → graph 1 (top, concave down, slope decreasing as t increases, so velocity decreasing)

- Object C: position increases at increasing rate (speeding up) → graph 3 (top, concave up, slope increasing as t increases, so velocity increasing)

But the bottom graph has graphs 4,5,6. Maybe the first three (1,2,3) are for positive motion, and 4,5,6 for negative? No, the dot diagrams have motion to the right. So:

A matches graph 2, B matches graph 1, C matches graph 3? But the problem says "Numbers can be used only one time" (1 - 6). Wait, maybe the bottom graph is also for positive motion but different accelerations. Wait, graph 4 (bottom) is concave up (slope increasing), graph 5 (bottom) is concave down (slope decreasing), graph 6 (bottom) is straight line (slope constant).

Ah! Maybe I made a mistake in the top graph's concavity. Let's look at the dot diagrams again:

- Object A: dots are equally spaced. So $\Delta x$ is constant for each $\Delta t$ → constant velocity → $x(t)=v_0t + x_0$, linear, so graph 6 (bottom, straight line) or graph 2 (top, straight line). Wait, the bottom graph's time arrow is to the left, so t increases to the left. So for graph 6 (bottom), as t increases (left), position decreases (negative slope), which is moving left. For graph 2 (top), t increases to the right, position increases (positive slope), moving right. So Object A (moving right, constant velocity) → graph 2.

- Object B: dots getting closer (in the direction of motion, right). So as t increases, $\Delta x$ decreases → velocity $v = \frac{\Delta x}{\Delta t}$ decreases. So position - time graph: $x(t)$ is a function where the slope (v) decreases. So concave down (since the second derivative is negative, acceleration negative). Graph 5 (bottom) is concave down (slope decreasing as t increases left), or graph 1 (top) is concave down (slope decreasing as t increases right). Let's take t increasing to the right (top graph). So Object B: moving right, slowing down → graph 1 (top, concave down, slope decreasing as t increases, so velocity decreasing).

- Object C: dots getting farther (in the direction of motion, right). So as t increases, $\Delta x$ increases → velocity $v=\frac{\Delta x}{\Delta t}$ increases. So position - time graph: concave up (second derivative positive, acceleration positive). Graph 3 (top) is concave up (slope increasing as t increases), or graph 4 (bottom) is concave up (slope increasing as t increases left). Let's take top graph: Object C → graph 3.

But the problem has numbers 1 - 6, and we have to use each number once. Wait, maybe the bottom graph is also part of the options. Let's re - assign:

- Object A: constant velocity → graph 6 (bottom, straight line, but t increases left, so if we consider t increasing right, maybe graph 2 is wrong. Wait, maybe the key is:

Dot spacing:

- A: equal spacing → constant velocity → linear graph (graph 2 or 6)

- B: spacing decreasing (closer) → velocity decreasing → graph with decreasing slope (graph 1 or 5)

- C: spacing increasing (farther) → velocity increasing → graph with increasing slope (graph 3 or 4)

Now, let's match:

- A: constant velocity → graph 2 (top, linear, positive slope) or graph 6 (bottom, linear, negative slope). Since motion is right, positive slope, so graph 2.

- B: velocity decreasing → graph 1 (top, concave down, slope decreasing) or graph 5 (bottom, concave down, slope decreasing). Let's take graph 1.

- C: velocity increasing → graph 3 (top, concave up, slope increasing) or graph 4 (bottom, concave up, slope increasing). Let's take graph 3.

But the problem says "Numbers can be used only one time", so 1,2,3 or 4,5,6? Wait, maybe the correct matching is:

- A: constant velocity → graph 2 (top)

- B: decelerating → graph 5 (bottom, concave down)

- C: accelerating → graph 4 (bottom, concave up)

No, this is getting confusing. Let's use the standard dot - diagram to position - time graph conversion:

- Constant velocity (equal spacing): linear position - time graph (straight line).

- Decelerating (spacing decreasing in direction of motion): position - time graph with decreasing slope (concave down).

- Accelerating (spacing increasing in direction of motion): position - time graph with increasing slope (concave up).

Now, looking at all 6 graphs:

- Linear graphs: 2 (top), 6 (bottom)

- Concave down (decreasing slope): 1 (top), 5 (bottom)

- Concave up (increasing slope): 3 (top), 4 (bottom)

Now, assign:

- A (constant velocity) → 2 (top, positive slope, moving right) or 6 (bottom, negative slope, moving left). Since motion is right, 2.

- B (decelerating, moving right) → 1 (top, concave down, slope decreasing as t increases, so velocity decreasing) or 5 (bottom, concave down, slope decreasing as t increases left, so velocity decreasing). Let's take 1.

- C (accelerating, moving right) → 3 (top, concave up, slope increasing as t increases, so velocity increasing) or 4 (bottom, concave up, slope increasing as t increases left, so velocity increasing). Let's take 3.

But the problem might have the bottom graph's time axis as standard (t right), but the arrow is to the left, which is non - standard. Alternatively, maybe the correct answer is:

A matches graph 2, B matches graph 5, C matches graph 4? No, I think the intended answer is:

- A: constant velocity → graph 2

- B: decelerating → graph 1

- C: accelerating → graph 3

But let's check again.

Wait, another approach:

- Object A: equal dots → constant speed → linear graph (graph 2 or 6). Let's say graph 2.

- Object B: dots getting closer → slowing down → graph with decreasing slope (graph 1 or 5). Let's say graph 1.

- Object C: dots getting farther → speeding up → graph with increasing slope (graph 3 or 4). Let's say graph 3.

But the problem says "Numbers can be used only one time", so 1,2,3.

Answer:

A matches graph: 2
B matches graph: 1
C matches graph: 3

(Note: If the bottom graph's time axis is considered, the answer might change, but based on standard position - time graph conventions with t increasing to the right, this is the most probable matching.)