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The speed at which Heather runs the \(100 \thinspace \rm{m}\) sprint is
consistent. The time it takes her to complete the race depends upon the
headwind she is working against. With a headwind of \(2 \thinspace
\rm{m/s}\), it takes Heather \(4\) seconds longer to finish the race
than if she has no headwind. To the nearest hundredth, determine
Heather’s average sprint speed, with the headwind, for the \(100
\thinspace \rm{m}\) sprint.
Let \(x\) represent Heather's speed with no headwind.
In this scenario, the headwind time is \(4\) seconds longer than without a headwind. Therefore, the equation is:
NPVs: \(x \ne 0, \thinspace 2\)
LCD: \(x(x - 2)\)
Because a negative speed does not make sense, it can be ignored.
Verify for \(x \doteq 8.14\).
Because approximate numbers are used, the answers are not exactly the same, but they are very close.
With no headwind, Heather’s average speed is \(8.14\thinspace \rm{m/s}\). Therefore, with a headwind, her average speed is \(6.14\thinspace \rm{m/s}\).
| Direction | Distance (km) | Speed (km/h) | Time |
| No Headwind | \(100\) |
\(x\)
|
\(\frac{100}{x}\)
|
|
Headwind of \(2 \thinspace \rm{m/s}\)
|
\(100\) |
\(x - 2\)
|
\(\frac{100}{x - 2}\) |
In this scenario, the headwind time is \(4\) seconds longer than without a headwind. Therefore, the equation is:
\[\frac{{100}}{x} + 4 = \frac{{100}}{{x - 2}}\]
NPVs: \(x \ne 0, \thinspace 2\)
LCD: \(x(x - 2)\)
\[\begin{align}
\left( {\frac{{100}}{{{\color{red}\cancel{\color{#444}{x}}}}}} \right)\left( {{\color{red}\cancel{\color{#444}{x}}}} \right)\left( {x - 2} \right) + \left( 4 \right)\left( x \right)\left( {x - 2} \right) &= \left( {\frac{{100}}{{{\color{red}\cancel{\color{#444}{x - 2}}}}}} \right)\left( x \right) {\color{red}\cancel{\color{#444}{\left( {x - 2} \right)}}} \\
100\left( {x - 2} \right) + 4x\left( {x - 2} \right) &= 100x \\
100x - 200 + 4x^2 - 8x &= 100x \\
4x^2 - 8x - 200 &= 0 \\
4\left( {x^2 - 2x - 50} \right) &= 0 \\
\end{align}\]
\left( {\frac{{100}}{{{\color{red}\cancel{\color{#444}{x}}}}}} \right)\left( {{\color{red}\cancel{\color{#444}{x}}}} \right)\left( {x - 2} \right) + \left( 4 \right)\left( x \right)\left( {x - 2} \right) &= \left( {\frac{{100}}{{{\color{red}\cancel{\color{#444}{x - 2}}}}}} \right)\left( x \right) {\color{red}\cancel{\color{#444}{\left( {x - 2} \right)}}} \\
100\left( {x - 2} \right) + 4x\left( {x - 2} \right) &= 100x \\
100x - 200 + 4x^2 - 8x &= 100x \\
4x^2 - 8x - 200 &= 0 \\
4\left( {x^2 - 2x - 50} \right) &= 0 \\
\end{align}\]
\[\begin{align}
a &= 1, \thinspace b = - 2, \thinspace c = - 50 \\
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
x &= \frac{{ - \left( { - 2} \right) \pm \sqrt {\left( { - 2} \right)^2 - 4\left( 1 \right)\left( { - 50} \right)} }}{{2\left( 1 \right)}} \\
x &= \frac{{2 \pm \sqrt {204} }}{2} \\
x &= \frac{{2 \pm 2\sqrt {51} }}{2} \\
x &= 1 \pm \sqrt {51} \\
x &\doteq 8.14 \thinspace {\rm{and}} \thinspace x \doteq - 6.14 \\
\end{align}\]
a &= 1, \thinspace b = - 2, \thinspace c = - 50 \\
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
x &= \frac{{ - \left( { - 2} \right) \pm \sqrt {\left( { - 2} \right)^2 - 4\left( 1 \right)\left( { - 50} \right)} }}{{2\left( 1 \right)}} \\
x &= \frac{{2 \pm \sqrt {204} }}{2} \\
x &= \frac{{2 \pm 2\sqrt {51} }}{2} \\
x &= 1 \pm \sqrt {51} \\
x &\doteq 8.14 \thinspace {\rm{and}} \thinspace x \doteq - 6.14 \\
\end{align}\]
Because a negative speed does not make sense, it can be ignored.
Verify for \(x \doteq 8.14\).
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r} \frac{{100}}{x} + 4 \\ \frac{{100}}{{8.14}} + 4 \\ 12.285... + 4 \\ 16.285... \\ \end{array}\] |
\[\begin{array}{l} \frac{{100}}{{x - 2}} \\ \frac{{100}}{{8.14 - 2}} \\ \frac{{100}}{{6.14}} \\ 16.286... \\ \end{array}\] |
| LS \(\doteq\) RS | |
Because approximate numbers are used, the answers are not exactly the same, but they are very close.
With no headwind, Heather’s average speed is \(8.14\thinspace \rm{m/s}\). Therefore, with a headwind, her average speed is \(6.14\thinspace \rm{m/s}\).
| For further information about problem solving with rational equations see pp. 345 to 347 of Pre-Calculus 11. |