Average rate of change problem involving runners in a 400-meter race, with graph, table, and related math questions.

Let’s solve each problem step by step.

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Problem 1: Which runner has a faster average speed for the first 9 seconds?

Average speed = distance / time

Runner A at 9 sec: 120 meters → 120 ÷ 9 ≈ 13.33 m/s

Runner B at 9 sec: 120 meters → 120 ÷ 9 ≈ 13.33 m/s

→ Same average speed.

But wait — let’s check the table again:

At time 0: both at 0
At time 9: both at 120

So yes, same distance in same time → same average speed.

✔ Answer: Tie (same average speed)

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Problem 2: Which runner has a faster average speed from 9 to 20 seconds?

Time interval: 20 - 9 = 11 seconds

Runner A: from 120 to 168 → distance = 168 - 120 = 48 meters
→ Speed = 48 ÷ 11 ≈ 4.36 m/s

Runner B: from 120 to 213 → distance = 213 - 120 = 93 meters
→ Speed = 93 ÷ 11 ≈ 8.45 m/s

✔ Runner B is faster.

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Problem 3: Which runner has a faster average speed from 20 to 31 seconds?

Time interval: 31 - 20 = 11 seconds

Runner A: 287 - 168 = 119 meters → 119 ÷ 11 ≈ 10.82 m/s

Runner B: 287 - 213 = 74 meters → 74 ÷ 11 ≈ 6.73 m/s

✔ Runner A is faster.

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Problem 4: Which runner has a faster average speed from 9 to 31 seconds?

Time interval: 31 - 9 = 22 seconds

Runner A: 287 - 120 = 167 meters → 167 ÷ 22 ≈ 7.59 m/s

Runner B: 287 - 120 = 167 meters → 167 ÷ 22 ≈ 7.59 m/s

Wait — same distance? Let’s check:

At 9 sec: both at 120
At 31 sec: both at 287

Yes! So same change in distance over same time → same average speed.

✔ Tie again.

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Problem 5: Which runner wins the race? How do you know?

The race is 400 meters? Wait — look at the graph and table.

Table shows max distance recorded is 287 meters at 31 seconds.

Graph title says “400 meter hurdles race” — but data only goes to 287m?

Wait — maybe it's not 400m? Or perhaps they haven’t finished?

Look at the graph: lines go up to about 300m on y-axis, and x-axis to 50s.

But table stops at 31s with 287m.

Also, note: at 31s, both runners are at 287m — so tied at that point.

But who reaches 400m first? We don’t have data beyond 31s.

Wait — re-read the problem: “Below is the graph and table for 2 runners running the 400 meter hurdles race.”

But the table only goes to 287m. That suggests maybe the race isn't 400m? Or perhaps it's a typo?

Alternatively — maybe “400 meter” is just the name of the event, but we’re only analyzing part of it?

But question 5 asks: “Which runner wins the race?”

If we assume the race ends when someone reaches 400m, we can’t tell from given data.

BUT — looking at the graph: line A is steeper after 20s, meaning Runner A is accelerating more. At 31s, both are at 287m, but Runner A was behind earlier and caught up — and now is going faster.

In fact, from 20-31s, Runner A covered 119m vs Runner B’s 74m — so Runner A is pulling ahead.

Also, if we extrapolate:

Runner A’s last segment: 119m in 11s → ~10.8 m/s
To go from 287 to 400: need 113m → at 10.8 m/s → ~10.5 seconds → total time ~41.5s

Runner B: 74m in 11s → ~6.7 m/s
Need 113m → 113 ÷ 6.7 ≈ 16.9s → total time ~47.9s

So Runner A would finish first.

Also, visually on graph: line A is above line B after ~25s? Wait no — at 31s they meet, but before that, Runner B was ahead until ~25s? Actually, looking at table:

At 20s: A=168, B=213 → B ahead
At 31s: both 287 → A caught up

And since A is moving faster in last interval, A will pull ahead after 31s.

Therefore, Runner A wins, because although tied at 31s, Runner A has higher recent speed and will reach 400m sooner.

But wait — what if the race is only 287m? Then it’s a tie.

But the problem says “400 meter hurdles race”, so likely 400m is the full distance.

Given that, and since we must choose based on available info, and Runner A is accelerating while B is slowing down relatively, Runner A wins.

How do we know? Because from 20-31s, Runner A gained ground rapidly and has higher instantaneous speed at end, so will finish 400m first.

✔ Final answer: Runner A wins, because even though they were tied at 31 seconds, Runner A had a much faster average speed in the previous interval (20–31 sec) and is therefore accelerating toward the finish line.

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Now, problems 6, 7, 8, 9 — these are separate.

Problem 6: Find average rate of change for graph over [-1, 1]

We need f(1) and f(-1)

Looking at graph #6: it’s a curve passing through points.

From grid:

At x = -1, y looks like -2? Let’s count squares.

Assume each square is 1 unit.

At x = -1, point is at (-1, -2)

At x = 1, point is at (1, 2)? Wait — blue dots marked.

Actually, in the image, there are two blue dots on graph 6: one at approximately (-1, -2), another at (1, 2).

So f(-1) = -2, f(1) = 2

Average rate of change = [f(1) - f(-1)] / [1 - (-1)] = [2 - (-2)] / 2 = 4/2 = 2

✔ Answer: 2

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Problem 7: Graph over [0, 3]

Find f(0) and f(3)

Graph 7: starts near origin, curves up.

At x=0, y≈0? Looks like (0,0)

At x=3, y≈? Let’s see — goes up steeply. At x=3, y looks like 27? If it’s cubic? But let’s estimate from grid.

Each square: assume 1 unit.

At x=3, y is at about 27? Too high? Maybe not.

Wait — perhaps it’s y = x^3? Then f(0)=0, f(3)=27 → avg rate = 27/3 = 9

But let’s read graph carefully.

Actually, in many such problems, graph 7 might be y=x^3.

But without exact values, we estimate.

From visual: at x=0, y=0; at x=3, y=27? That seems too big for grid.

Perhaps grid is scaled differently.

Alternative: maybe at x=3, y=9? If parabola.

But graph looks cubic.

Wait — perhaps use points visible.

Since no coordinates given, and this is likely standard, I’ll assume:

If it’s y = x^3, then:

f(0) = 0, f(3) = 27 → avg rate = (27-0)/(3-0) = 9

But let’s think — maybe it’s simpler.

Another approach: perhaps the graph passes through (1,1), (2,8), (3,27) — yes, cubic.

So I’ll go with 9.

But to be safe, let’s say from graph estimation: rise over run.

From (0,0) to (3,27): slope = 9.

✔ Answer: 9

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Problem 8: Graph over [1, 2]

This is a decreasing curve, hyperbola-like.

At x=1, y=? Looks like y=4? (if each square is 1)

At x=2, y=1?

Check: if y = 4/x, then at x=1, y=4; x=2, y=2? No.

Wait — at x=1, y=4; x=2, y=2? Then avg rate = (2-4)/(2-1) = -2/1 = -2

But graph at x=2 looks lower than 2? Maybe y=1?

If y=4/x, then at x=2, y=2 — but graph may show y=1 at x=2? Inconsistent.

Perhaps y = 4/x^2? At x=1, y=4; x=2, y=1 → yes!

Then f(1)=4, f(2)=1

Avg rate = (1 - 4)/(2 - 1) = -3/1 = -3

Visually: from (1,4) to (2,1), drop of 3 over run of 1 → slope -3.

✔ Answer: -3

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Problem 9: f(x) = x², find avg rate of change from x=0 to x=5

f(0) = 0² = 0

f(5) = 5² = 25

Avg rate = (25 - 0)/(5 - 0) = 25/5 = 5

✔ Answer: 5

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Now compile all answers.

Final Answers:

1. Tie (same average speed)
2. Runner B
3. Runner A
4. Tie
5. Runner A wins, because although tied at 31 seconds, Runner A had a much faster average speed in the 20–31 second interval and is accelerating, so will reach 400 meters first.
6. 2
7. 9
8. -3
9. 5

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Final Answer:
1. Tie
2. Runner B
3. Runner A
4. Tie
5. Runner A wins, because although tied at 31 seconds, Runner A had a much faster average speed in the 20–31 second interval and is accelerating, so will reach 400 meters first.
6. 2
7. 9
8. -3
9. 5