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AfilCa [17]
3 years ago
9

Explain why if a runner completes a 6.2â-mi race in 32 âmin, then he must have been running at exactly 11 âmi/hr at least twice

in the race. Assume theâ runner's speed at the finish line is zero.
Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

Answer:

Accelerating to top speed, deaccelerating to finish line.

Step-by-step explanation:

If the runner kept a constant speed of 11 mph for the whole duration of his run (32 minutes), the distance he would have covered is:

d=11*\frac{32}{60}\\d= 5.87\ mi

This means that, in order to run the full 6.2 miles, the runner needs to reach a speed over 11 mph. Assume he starts from rest, while accelerating the runner reaches, and the surpasses, the 11 mph mark. Since his speed at the finish line is zero, the runner has to deaccelerate from his current running speed (which should be higher than 11 mph), passing through 11 mph and reaching zero at the finish line.

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Pie

\bold{\huge{\underline{ Solution }}}

<u>We </u><u>have</u><u>, </u>

  • Line segment AB
  • The coordinates of the midpoint of line segment AB is ( -8 , 8 )
  • Coordinates of one of the end point of the line segment is (-2,20)

Let the coordinates of the end point of the line segment AB be ( x1 , y1 ) and (x2 , y2)

<u>Also</u><u>, </u>

Let the coordinates of midpoint of the line segment AB be ( x, y)

<u>We </u><u>know </u><u>that</u><u>, </u>

For finding the midpoints of line segment we use formula :-

\bold{\purple{ M( x,  y) = }}{\bold{\purple{\dfrac{(x1 +x2)}{2}}}}{\bold{\purple{,}}}{\bold{\purple{\dfrac{(y1 + y2)}{2}}}}

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

  • The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .

<u>For </u><u>x </u><u>coordinates </u><u>:</u><u>-</u>

\sf{  -8  = }{\sf{\dfrac{(- 2 +x2)}{2}}}

\sf{2}{\sf{\times{ -8  = - 2 + x2 }}}

\sf{ - 16 = - 2 + x2 }

\sf{ x2 = -16 + 2 }

\bold{ x2 = -14  }

<h3><u>Now</u><u>, </u></h3>

<u>For </u><u>y </u><u>coordinates </u><u>:</u><u>-</u>

\sf{  8  = }{\sf{\dfrac{(- 20 +x2)}{2}}}

\sf{2}{\sf{\times{ 8   = - 20 + x2 }}}

\sf{ 16 = - 20 + x2 }

\sf{ y2 = 16 + 20 }

\bold{ y2 = 36  }

Thus, The coordinates of another end points of line segment AB is ( -14 , 36)

Hence, Option A is correct answer

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