Solution 10.4

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(New page: We first calculate the particle´s acceleration using Newton´s Second Law <math>F=ma</math>. <math>\begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-...)
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(New page: We first calculate the particle´s acceleration using Newton´s Second Law <math>F=ma</math>. <math>\begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-...)
 

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We first calculate the particle´s acceleration using Newton´s Second Law \displaystyle F=ma.


\displaystyle \begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-2}} \\ \end{align}

Now using the kinematic equation (see section 6),

\displaystyle v=u+at\ gives

\displaystyle \begin{align} & 10=0+2\textrm{.}5\times t \\ & \\ & t=4\ \text{s} \\ \end{align}