To determine when bikes A and B will be the same distance from the starting point, we can set up equations for their distances from the starting point over time.
Let \(d_a\) be the distance of bike A from the starting point and \(d_b\) be the distance of bike B from the starting point.
-
Bike A is currently \(50\) miles from the starting point and travels at \(30\) mph. Therefore, after \(t\) hours, the distance of bike A from the starting point can be expressed as: \[ d_a = 50 - 30t \] (since bike A is moving towards the starting point).
-
Bike B is currently \(100\) miles from the starting point and travels at \(25\) mph. Therefore, after \(t\) hours, the distance of bike B from the starting point can be expressed as: \[ d_b = 100 - 25t \] (since bike B is also moving towards the starting point).
To find out when they will be the same distance from the starting point, we set \(d_a\) equal to \(d_b\): \[ 50 - 30t = 100 - 25t \]
Now, let's solve for \(t\):
- Rearranging the equation gives us: \[ 30t - 25t = 100 - 50 \] \[ 5t = 50 \] \[ t = 10 \]
Now that we have \(t\), we can calculate the distance each bike has traveled and the distance from the starting point.
For bike A: \[ d_a = 50 - 30(10) = 50 - 300 = -250 \text{ miles} \]
Since negative distance doesn't make sense in this context, we realize bike A will reach the starting point before bike B meets it, meaning we should simply find out the distance at the moment of equality.
To find out how far each bike is from the starting point after 10 hours:
- Bike A would have actually reached the starting point way before \(t = 10\) hours.
- Initalizing time from when both bikes start approaching and finding when they’ll meet gives: At \(t=0\), at their starting point respectively; as they meet at the 100 mile home stretch:
- Bike A after travelling \(50\) miles.
- Bike B after travelling \(100\) miles when they meet.
From these distances, we analyze where \(d_a < d_b\) till they meets hence both bikes will have traveled with respect to starting points equating the respective ride routes.
We conclude that there are no scenarios where they would reach the same distance from the start as Bike A would terminate travel before Bike B could yield equality at 50 miles.
Ultimately, they're simply following their travel paths stating A (30 mph) will reach home in \( \frac{50}{30} \) \(= 1.67 \text{ hours}\). Thus, no "same distance" dynamic takes place on equivalent travel as stated.
So, Bike A and B will be the same distance from the starting point when they meet actually at 50-mile mark before concluding the loop prior. In other word’s, they meet at \(50\) miles having computed their respective inputs further explicated as required thereby yielding the same distance at their respective routes calculated throughout their journeys across stipulated mile timed towards the point again reiterating, they will not converge at a foreseeable mile together.