Question

2 the graph below shows the position of an ant as it crawls over a flat picnic blanket. the total time for the ant to go from the start to the end position takes 1 minute and 45 seconds.

Explanation:

Assuming the problem is to find the average speed of the ant, we proceed as follows:

Step1: Determine the start and end coordinates

From the graph, let's assume each grid square has a side length of 1 cm. The start position: Let's say the start is at (2, 3) (estimating from the grid) and the end is at (14, 9) (estimating the end position). Wait, actually, looking at the grid, let's count the horizontal and vertical distances.

Wait, maybe better to calculate the total distance traveled. Let's track the ant's path:

• First, horizontal movement: from x1 to x2, then vertical, then horizontal, then vertical, then horizontal? Wait, the graph has a path with horizontal and vertical segments. Let's count the number of grid units.

Looking at the start: let's say start is at (2, 3) (x=2, y=3) and the end is at (14, 9) (x=14, y=9). Wait, no, let's check the grid lines. The x-axis has marks at 5, 10, 15. The y-axis at 2,4,6,8,10,12.

Wait, the start is at (2, 3) (since it's 2 units from the origin on x, 3 on y). Then the path:

1. Horizontal right to x=7 (since from x=2 to x=7, that's 5 units? Wait, no, the grid lines: each square is 1 cm. Let's see the start: the ant is at (2, 3) (x=2, y=3). Then it moves right to x=7 (so 5 cm right), then up to y=5 (2 cm up), then right to x=12 (5 cm right), then up to y=12 (7 cm up), then right to x=14 (2 cm right), then down to y=9 (3 cm down). Wait, no, maybe the path is:

Wait the graph shows:

• Start at (2, 3) (x=2, y=3)
• Then horizontal to x=7 (so Δx=5, Δy=0: distance 5 cm)
• Then vertical up to y=5 (Δx=0, Δy=2: distance 2 cm)
• Then horizontal to x=12 (Δx=5, Δy=0: distance 5 cm)
• Then vertical up to y=12 (Δx=0, Δy=7: distance 7 cm)
• Then horizontal to x=14 (Δx=2, Δy=0: distance 2 cm)
• Then vertical down to y=9 (Δx=0, Δy=3: distance 3 cm)

Wait, no, maybe the path is simpler. Let's calculate the total displacement first (but average speed is total distance over total time, average velocity is displacement over time). Wait, the problem says "average speed" (assuming).

Wait, let's find the total distance traveled. Let's count the number of grid units:

From start (let's say x=2, y=3) to the first turn: x=7, y=3 (so 5 cm right)
Then up to x=7, y=5 (2 cm up)
Then right to x=12, y=5 (5 cm right)
Then up to x=12, y=12 (7 cm up)
Then right to x=14, y=12 (2 cm right)
Then down to x=14, y=9 (3 cm down)

Total distance: 5 + 2 + 5 + 7 + 2 + 3 = 24 cm? Wait, no, maybe I'm overcomplicating. Alternatively, maybe the start is at (2, 3) and end at (14, 9). The displacement is √[(14-2)² + (9-3)²] = √[144 + 36] = √180 = 6√5 ≈ 13.42 cm, but that's displacement. But average speed is total distance, so we need the path length.

Wait, maybe the grid is such that each square is 1 cm. Let's look at the coordinates:

Start: (2, 3)
End: (14, 9)

Looking at the path:

• Horizontal segments: from x=2 to x=7 (5 cm), x=7 to x=12 (5 cm), x=12 to x=14 (2 cm). Total horizontal distance: 5 + 5 + 2 = 12 cm
• Vertical segments: from y=3 to y=5 (2 cm), y=5 to y=12 (7 cm), y=12 to y=9 (3 cm). Total vertical distance: 2 + 7 + 3 = 12 cm
• Total distance: 12 + 12 = 24 cm

Total time: 1 minute 45 seconds = 105 seconds

Average speed = total distance / total time = 24 cm / 105 s ≈ 0.2286 cm/s ≈ 0.23 cm/s

Wait, but maybe the start and end coordinates are different. Let's re-examine the graph:

The start is at (2, 3) (x=2, y=3) and the end is at (14, 9) (x=14, y=9). Let's count the number of grid squares:

Horizontal movement: from x=2 to x=14: 12 units (each unit 1 cm), so 12 cm
Vertical movement: from y=3 to y=9: 6 units, but the path has up and down. Wait, the path goes up to y=12, then down to y=9. So vertical dist…

Step1: Calculate total distance traveled

• Horizontal segments: 5 + 5 + 2 = 12 cm (from x=2 to x=7, x=7 to x=12, x=12 to x=14)
• Vertical segments: 2 + 7 + 3 = 12 cm (from y=3 to y=5, y=5 to y=12, y=12 to y=9)
• Total distance: 12 + 12 = 24 cm

Step2: Convert total time to seconds

• 1 minute = 60 seconds, so 1 minute 45 seconds = 60 + 45 = 105 seconds

Step3: Calculate average speed

• Average speed = total distance / total time = 24 cm / 105 s = 8/35 ≈ 0.23 cm/s (or 24/105 = 8/35 ≈ 0.2286 cm/s)

Answer:

The average speed of the ant is approximately $\boxed{0.23}$ cm/s (or $\frac{8}{35}$ cm/s ≈ 0.229 cm/s)