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jarptica [38.1K]
4 years ago
15

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 init

ially 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).
Physics
1 answer:
sergeinik [125]4 years ago
4 0

Answer:

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.

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