The first tangential acceleration $a_{1T}$ is given by the formula $v_{1}^{2}/2s_{1}$, where $v_{1}$ is the initial speed (40 km/h converted to m/s by dividing by 3.6) and $s_{1}$ is the distance (60 m). So, we have:
\[a_{1T} = \frac{{(40/3.6)^{2}}}{{2 \times 60}}
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