Solution 6.8b

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We investigate the distance after \displaystyle 10\ \text{s}.

Using \displaystyle s=ut+\frac{1}{2}a{{t}^{\ 2}} gives


\displaystyle s=0+\frac{1}{2}\times \left( 0 \textrm{.}075 \right)\times {{10}^{2}}=37 \textrm{.}5\ \text{m}