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If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runner A has a 55.0-meter head start?

Answer :

Runner B will catch Runner A after approximately 44.00 seconds.

To determine the time it takes for Runner B to catch Runner A, we need to consider their relative speeds and the initial distance between them.

Runner B is running faster than Runner A, so the distance between them will decrease over time. The relative speed of Runner B with respect to Runner A is the difference in their speeds:

Relative speed = Speed of Runner B - Speed of Runner A

Relative speed = 7.90 m/s - 7.50 m/s = 0.40 m/s.

Runner B needs to cover the initial distance (head start) of 55.0 meters between them. The time it takes for Runner B to catch up to Runner A can be calculated using the formula:

Time = Distance / Relative Speed.

Time = 55.0 m / 0.40 m/s = 137.5 seconds.

However, this time includes the time taken by Runner A to run the initial distance. To find the time taken by Runner B to catch Runner A after accounting for the head start, we subtract the initial time from the total time:

Time taken by Runner B = Total time - Time taken by Runner A.

Runner A's initial time = Distance / Speed = 55.0 m / 7.50 m/s = 7.3333 seconds.

Time taken by Runner B = 137.5 seconds - 7.3333 seconds = 130.1667 seconds.

Rounded to two decimal places, it takes Runner B approximately 130.17 seconds to catch up to Runner A.

To learn more about head start, here

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Rewritten by : Barada

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

Slopes:

Runner A: y = 7.50x + 55

Runner B: y = 7.90 x

sooooo

7.50 x + 55 = 7.90 x

- 7.50 x - 7.50 x

55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 = 1086.25 m