To find the distance the faster motorcyclist travels to overtake the slower one, we need to determine their speeds and the time it takes to close the gap.
The speeds are in the ratio $5:3$. The slower motorcyclist's speed ($S_{slow}$) is 12 km/h. Let the faster motorcyclist's speed be $S_{fast}$.
We set up the proportion:
$ \frac{S_{fast}}{S_{slow}} = \frac{5}{3} $Substitute the known speed:
$ \frac{S_{fast}}{12 \text{ km/h}} = \frac{5}{3} $Solve for $S_{fast}$:
$ S_{fast} = \frac{5}{3} \times 12 \text{ km/h} $ $ S_{fast} = 5 \times 4 \text{ km/h} $ $ S_{fast} = 20 \text{ km/h} $The faster motorcyclist travels at 20 km/h.
The slower motorcyclist has a 24 km head start. The faster motorcyclist needs to close this distance.
The relative speed at which the faster motorcyclist catches up is the difference between their speeds:
$ S_{relative} = S_{fast} - S_{slow} $ $ S_{relative} = 20 \text{ km/h} - 12 \text{ km/h} $ $ S_{relative} = 8 \text{ km/h} $The time required to cover the 24 km gap is calculated using:
$ \text{Time} = \frac{\text{Distance Gap}}{S_{relative}} $ $ \text{Time} = \frac{24 \text{ km}}{8 \text{ km/h}} $ $ \text{Time} = 3 \text{ hours} $It will take 3 hours for the faster motorcyclist to overtake the slower one.
To find how far the faster motorcyclist travels, multiply their speed by the time taken to overtake:
$ \text{Distance} = S_{fast} \times \text{Time} $ $ \text{Distance} = 20 \text{ km/h} \times 3 \text{ hours} $ $ \text{Distance} = 60 \text{ km} $The faster motorcyclist travels 60 km to overtake the slower motorcyclist.
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