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In a biathalon race you first ride a bicycle at an average speed of 21.8 mi/h for 16.5 miles, then you must run for another 5.5 miles. With what average speed, in miles per hour, must you run if your average speed for the entire race is to be 13.8 mi/h?

Respuesta :

Answer:

6.6 mi/h

Step-by-step explanation:

Given,

For first 16.5 miles, speed is 21.8 miles per hour,

Since,

[tex]Time = \frac{Distance}{Speed}[/tex]

So, the time taken in first time = [tex]\frac{16.5}{21.8}[/tex]

Let x be the average speed in miles per hour for another 5.5 miles,

So, the time taken in second time = [tex]\frac{5.5}{x}\text{ hours}[/tex]

∴ Total time taken in entire journey = [tex]\frac{16.5}{21.8}+\frac{5.5}{x}[/tex]

Now, total distance covered = 16.5 + 5.5 = 22 miles,

Hence, the average speed for the entire race = [tex]\frac{22}{\frac{16.5}{21.8}+\frac{5.5}{x}}[/tex]

According to the question,

[tex]\frac{22}{\frac{16.5}{21.8}+\frac{5.5}{x}}=13.8[/tex]

[tex]22=13.8(\frac{16.5}{21.8}+\frac{5.5}{x})[/tex]

[tex]22=\frac{227.7}{21.8}+\frac{75.9}{x}[/tex]

[tex]22-\frac{117.7}{21.8}=\frac{75.9}{x}[/tex]

[tex]\frac{361.9}{21.8}=\frac{75.9}{x}[/tex]

[tex]\implies x = 6.56856\approx 6.6\text{ miles per hour}[/tex]

onyebuchinnaji

The average speed for this additional distance is 6.57 mi/h

The given parameters include;

  • the average speed on bicycle = 21.8 mi/h
  • the distance traveled on bicycle = 16.5 miles
  • additional distance to be covered = 5.5 miles
  • let the average speed for this additional distance = v
  • the average speed for the entire race = 13.8 mi/h

To find:

  • the average speed for this additional distance,  v

The average speed formula is given as;

[tex]average\ speed = \frac{total \ distance }{total \ time} \\\\average\ speed = \frac{16.5 \ miles \ +\ 5.5 \ miles}{\frac{16.5 \ mi}{21.8 \ mi/h}\ \ +\ \ \frac{5.5 \ mi}{v} } \\\\13.8 \ mi/h = \frac{(16.5 \ miles \ +\ 5.5 \ miles)}{(\frac{16.5 \ mi}{21.8 \ mi/h}\ \ +\ \ \frac{5.5 \ mi}{v} ) }\\\\13.8 = \frac{(16.5 \ +\ 5.5 )}{(\frac{16.5 }{21.8 }\ \ +\ \ \frac{5.5 }{v} ) }\\\\13.8 = \frac{22 }{0.757\ \ +\ \ \frac{5.5 }{v} }\\\\13.8 (0.757\ \ +\ \ \frac{5.5 }{v} ) = 22\\\\[/tex]

[tex]10.446 + \frac{75.9}{v} = 22\\\\\frac{75.9}{v} = 22-10.446\\\\\frac{75.9}{v} = 11.554\\\\v = \frac{75.9}{11.554} \\\\v = 6.57 \ mi/h[/tex]

Thus, the average speed for this additional distance is 6.57 mi/h

Learn more here: https://brainly.com/question/17289046

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