Runner a is initially 3.91 mi west of a flagpole and is running with a constant velocity of 6.04 mi/h due east. runner b is initially 2.95 mi east of the flagpole and is running with a constant velocity of 5.00 mi/h due west. how far are the runners from the flagpole when they meet? (west is positive and east is negative).

The runners are 0.159 mi to the west of the flagpole.

Explanation:

Let´s place the origin of the frame of reference at the point where the flagpole is located.

The initial position of runner A is 3.91 mi and his velocity is -6.04 mi/h

The initial position of runner B is -2.95 mi and his velocity is  5.00 mi/h

When both runners meet, their position is the same. The equation of the position of each runner is:

x = x0 + v · t

Where

x = position at time t

x0 = initial position

v = velocity

t = time

Then, at the meeting point:

x runner A = x runner B

3.91 mi - 6.04 mi/h · t = -2.95 mi + 5.00m/h · t

Solving for t:

3.91 mi + 2.95 mi =  5.00m/h · t + 6.04 mi/h · t

6.86 mi = 11.04 mi/h · t

t = 0.621 h

Now, we use this time to find the meeting point. We can use the equation of any runner. Let´s use the position of runner A:

x = 3.91 mi - 6.04 mi/h · 0.621 h = 0.159 mi

Since the position is positive, the runners met 0.159 mi to the west of the flagpole.

gets less

explanation:

the masses of the objects and the distance between them