Math worksheet solving a fox and rabbit chase problem, showing time, distance, and catch-up calculation.

Let's carefully analyze the problem and solution provided in the image, step by step.

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Problem Summary:

- A fox named Fred spots a rabbit named Roger who is 18 meters away.
- Both start running at the same time:
- Fred (fox) runs at 5.5 m/s.
- Roger (rabbit) runs at 3.5 m/s.
- Roger runs away from Fred in a straight line.
- The question is: How long will it take for Fred to catch up with Roger?
- Additionally, Roger’s hole is 35 meters from him when he sees the fox — so we need to determine if Roger can reach his hole before being caught.

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Step 1: Understanding the Motion

This is a relative motion problem. Since both are moving in the same direction (Fred chasing Roger), we can find how fast Fred is gaining on Roger.

#### Speed of Fred:
$ 5.5 $ m/s

#### Speed of Roger:
$ 3.5 $ m/s

#### Relative speed (how fast Fred closes the gap):
$$
5.5 - 3.5 = 2.0 \text{ m/s}
$$

So, Fred gains 2 meters every second on Roger.

#### Initial distance between them:
18 meters

#### Time to catch up:
$$
\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{18}{2} = 9 \text{ seconds}
$$

✔ So, Fred catches Roger after 9 seconds.

This matches the answer given in the image.

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Step 2: Check the Table

The table shows distances covered by both animals at each second:

| Time (s) | Fred Distance (m) | Roger Distance (m) |
|---------|-------------------|--------------------|
| 0 | 0 | 18 |
| 1 | 5.5 | 21.5 |
| 2 | 11 | 25 |
| 3 | 16.5 | 28.5 |
| 4 | 22 | 32 |
| 5 | 27.5 | 35.5 |
| 6 | 33 | 39 |
| 7 | 38.5 | 42.5 |
| 8 | 44 | 46 |
| 9 | 49.5 | 49.5 |
| 10 | 55 | 53 |

At t = 9 seconds, both have traveled:
- Fred: $ 5.5 \times 9 = 49.5 $ m
- Roger: started 18 m ahead → $ 18 + (3.5 \times 9) = 18 + 31.5 = 49.5 $ m

They meet at 49.5 meters from Fred’s starting point. ✔

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Step 3: Will Roger Make It to His Hole?

Roger’s hole is 35 meters from him when he sees the fox.

So, total distance from Fred to the hole = $ 18 + 35 = 53 $ meters.

We want to know: Can Roger reach 53 meters from Fred’s starting point before Fred reaches there?

Let’s calculate:

#### Time for Roger to reach the hole:
Roger needs to run 35 meters at 3.5 m/s:
$$
\text{Time} = \frac{35}{3.5} = 10 \text{ seconds}
$$

#### Where is Fred after 10 seconds?
Fred runs at 5.5 m/s:
$$
\text{Distance} = 5.5 \times 10 = 55 \text{ meters}
$$

So, after 10 seconds:
- Roger has reached 53 meters (the hole)
- Fred has reached 55 meters → already past the hole

But wait — does that mean Fred has already caught Roger?

Let’s check their positions at t = 10 s:
- Roger: $ 18 + 3.5 \times 10 = 18 + 35 = 53 $ m
- Fred: $ 5.5 \times 10 = 55 $ m

But they met at 49.5 m at t = 9 s, so Roger was caught before reaching the hole.

So even though Roger would reach the hole in 10 seconds, he is caught at 9 seconds, which is before he gets to the hole.

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✔ Final Answers:

1. It will take Fred 9 seconds to catch up with Roger.
→ Supported by relative speed calculation and table data.

2. Will Roger make it to his hole before being caught?
✘ No.
- Roger reaches the hole in 10 seconds, but
- Fred catches him at 9 seconds, so Roger does not make it.

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✔ Explanation from the Image (Verified):

> "If the hole is 35m away from him, then it is a total distance of 53m away from the fox. Roger will reach 53m in 10 seconds. By this time, Fred will have run 55m and overtaken him, so Roger will have been caught!"

✔️ This is correct — although it slightly misstates the timeline (Roger is caught at 9s, not 10s), the conclusion is right: Roger is caught before reaching the hole.

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🟩 Conclusion:

- Catch-up time: 9 seconds ✔
- Roger cannot escape — he is caught before reaching his hole ✔

The solution is correct and well-supported by both algebra and the table.