To solve this problem, we calculate whether the dog catches the cat before the cat reaches the oak tree.

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Step 1: Define the scenario mathematically.

• Dog's speed: $v_d = 15 \, \text{m/s}$
• Cat's speed: $v_c = 13 \, \text{m/s}$
• Initial distance between dog and cat: $d_i = 18 \, \text{m}$
• Distance of cat from the oak tree: $d_o = 79 \, \text{m}$

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Step 2: Determine the time it takes for the cat to reach the oak tree.

The time for the cat to reach the oak tree is calculated as: $t_c = \frac{d_o}{v_c}$ Substitute the values: $t_c = \frac{79}{13} \approx 6.08 \, \text{seconds}$

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Step 3: Determine the time for the dog to catch the cat.

The relative speed between the dog and the cat is: $v_{\text{relative}} = v_d - v_c = 15 - 13 = 2 \, \text{m/s}$ The time for the dog to close the 18 m gap is: $t_d = \frac{d_i}{v_{\text{relative}}}$ Substitute the values: $t_d = \frac{18}{2} = 9 \, \text{seconds}$

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Step 4: Compare the times.

• The cat reaches the oak tree in $t_c \approx 6.08 \, \text{seconds}.$
• The dog requires $t_d = 9 \, \text{seconds}$ to catch the cat.

Since $t_c < t_d$, the cat reaches the oak tree before the dog catches it.

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Conclusion:

The cat reaches the oak tree safely, and the dog does not catch the cat.

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Would you like further details or have any questions? 😊
Here are 5 related questions for further exploration:

1. What would happen if the initial gap was smaller?
2. How would the outcome change if the dog’s speed increased by 1 m/s?
3. How far from the oak tree would the dog catch the cat if the chase continued?
4. Can we generalize the result for any initial distance and speeds?
5. What is the distance covered by both animals before the cat reaches the oak tree?

Tip: Always analyze the relative speed when two objects are moving in the same direction to determine outcomes.