Let's define the distance from the starting point for Bike A and Bike B.
- Bike A is currently 50 miles from the starting point and travels at 30 miles per hour.
- Bike B is currently 100 miles from the starting point and travels at 25 miles per hour.
Let \( x \) be the number of hours after the start of the race.
The distance of Bike A from the starting point after \( x \) hours will be: \[ \text{Distance of Bike A} = 50 - 30x \]
The distance of Bike B from the starting point after \( x \) hours will be: \[ \text{Distance of Bike B} = 100 - 25x \]
We want to find when both bikes are the same distance from the starting point, so we set the two distances equal to each other: \[ 50 - 30x = 100 - 25x \]
Now, let's solve for \( x \):
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Rearranging the equation gives: \[ 50 - 100 = -25x + 30x \] \[ -50 = 5x \]
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Dividing both sides by 5: \[ x = -10 \]
This negative value indicates that Bike A was 10 miles further from the starting point than Bike B at the starting time.
To find the distance from the starting point when they are equal:
- After 10 hours (from the point when bike A left the starting line at 30 mph, covering the distance it travels):
For Bike A: \[ \text{Distance of Bike A} = 50 - 30 \times 10 = 50 - 300 = -250 , \text{(beyond finish line, hence not relevant)} \]
For Bike B: \[ \text{Distance of Bike B} = 100 - 25 \times 10 = 100 - 250 = -150 , \text{(also beyond finish line, not relevant)} \]
The key here is to realize what this signifies: In reality, we want the distances to equal.
To find how many miles until they are equal, we can set the future distances equal: Given that they start at their respective positions and we need an equation.
The distance each travels until they meet:
- If Bike A covers its distance to the point they meet: Let \( d \) be the distance traveled by Bike A until they meet and \( d/30 \) is the time taken as A goes towards, we adjust Bike B from behind: So, total distance from starting would define condition where they close the gap: Since A needs distance \( 50 - d \) Bike B needs to journey from 100 - d
So once we fix joint distance as Before on gap closes based on speeds, provide clarity in terms for both:
They Cover in relation: \[ 50 - 30t + 100 - 25t = 100-100 \] Solving with future values would provide gaps,
Here, meeting occurs at 50, as 80, so finally resolving you could see they eventually meet before traversing the direct lines towards meeting in simpler ratios.
Instead it pays more to clarify distance until they reach each other than those results summarily.
Conclusively, they meet when the values balance past any overlooked axiomatic values.
Thus, the distance you can provide thoroughly through situational resolution cited best distance directly is, covering speeds aligned to each other and time across with their travel distance when extensive coverage meets.