Question
A car starting from rest, accelerates uniformly with $5m/s^2$ for sometime and then decelerates to come to rest with $3m/s^2$. Find the maximum velocity attained during the motion and the distance covered in a total time of 6 seconds of the journey.

Answer

During acceleration, $\text{v}_\text{m}=0+5\times\text{t}_\text{a}$
$\Rightarrow \text{t}_\text{a}=\frac{\text{v}_\text{m}}{5}$
$\text{v}^2_\text{m}=0+2\times5\times\text{s}_\text{a}$
$\Rightarrow \text{s}_\text{a}=\frac{\text{v}^2_\text{m}}{10}$

During deceleration, $0=\text{v}_\text{m}-3\text{t}_\text{d}$
$\Rightarrow \text{t}_\text{d}=\frac{\text{v}_\text{m}}{3}$
$0=\text{v}^2_\text{m}-2\times3\times\text{s}_\text{d}$
$\Rightarrow \text{s}_\text{d}=\frac{\text{v}^2_\text{m}}{6}$ Total time $=6=\text{t}_\text{a}+\text{t}_\text{d}$
$\therefore \text{v}_\text{m}=\frac{6\times5\times3}{8}=11.25\text{ms}^{-1}$ Total length covered $=\text{s}_\text{a}+\text{s}_\text{d}=\text{v}^2_\text{m}\Big(\frac{1}{10}+\frac{1}{6}\Big)$
$=(11.25)^2\frac{16}{60}$
$=33.75\text{m}$

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