From Mechanics

Revision as of 12:30, 10 April 2010 (edit) Ian (Talk | contribs) (New page: We use the results of the first two parts of the question. As the sprinter runs <math>\text{16 m}</math> in the first stage of the race he must run <math>\text{84 m}</math> in the secon...) ← Previous diff		Current revision (12:38, 10 April 2010) (edit) (undo) Ian (Talk | contribs)
Line 7:		Line 7:
	in the second stage.		in the second stage.
-	For the second stage we use	+	For the second stage we use <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> with <math>a=0</math> giving
		+
		+	<math>\begin{align}
		+	& 84=8t+0 \\
		+	& t=10\textrm{.}5\ \text{s} \\
		+	\end{align}</math>
		+
		+
		+	The total time taken is
		+	<math>14\textrm{.}5\ \text{s}</math>

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Current revision

We use the results of the first two parts of the question.

As the sprinter runs \displaystyle \text{16 m} in the first stage of the race he must run \displaystyle \text{84 m} in the second stage.

For the second stage we use \displaystyle s=ut+\frac{1}{2}a{{t}^{\ 2}} with \displaystyle a=0 giving

\displaystyle \begin{align} & 84=8t+0 \\ & t=10\textrm{.}5\ \text{s} \\ \end{align}

The total time taken is \displaystyle 14\textrm{.}5\ \text{s}