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A father and his son went hiking in Tagaytay City on a specific
route. The father hiked at an average rate of 5 kph and the
at 6 kph. If the son began hiking 45 min after the father started
walking, how long did it take the son to overtake his father?.

Sagot :

sethartiolamarcelo

Answer:

Let's solve this step by step:

1. **Determine the head start of the father**: The father hiked for 45 minutes (which is 0.75 hours) before the son started. At 5 kph, he covered a distance of \( 5 \text{ kph} \times 0.75 \text{ hours} = 3.75 \text{ km} \).

2. **Set up the relative speed**: The son's speed is 6 kph, and the father's speed is 5 kph. The relative speed of the son with respect to the father is \( 6 \text{ kph} - 5 \text{ kph} = 1 \text{ kph} \).

3. **Calculate the time to overtake**: The son needs to cover the 3.75 km head start at the relative speed of 1 kph. The time \( t \) it takes for the son to overtake the father is given by:

\[ \text{time} = \frac{\text{distance}}{\text{relative speed}} = \frac{3.75 \text{ km}}{1 \text{ kph}} = 3.75 \text{ hours} \]

4. **Interpret the result**: It takes the son 3.75 hours to overtake his father after he starts hiking. This means that from the moment the son begins his hike, he will take 3.75 hours to catch up with his father.

5. **Verify the distances**: To confirm, in 3.75 hours, the father would have walked \( 3.75 \text{ hours} \times 5 \text{ kph} = 18.75 \text{ km} \) since his start. The son, starting 45 minutes later, would cover \( 3.75 \text{ hours} \times 6 \text{ kph} = 22.5 \text{ km} \), overtaking his father precisely at the 22.5 km mark on the route.

Thus, it takes the son 3.75 hours to overtake his father after he starts hiking.

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