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A body moves with a velocity of $2\, m s ^{-1}$ for $5\, s$, then its velocity increases uniformly to $10\, m s ^{-1}$ in next $5\, s.$ Thereafter, its velocity begins to decrease at a uniform rate until it comes to rest after $5\, s$.
$(i)$ Plot a velocity-time graph for the motion of the body.
$(ii)$ From the graph, find the total distance covered by the body after $2\, s$ and $12\, s$.
Solution
$(ii)$ According to the graph :
Distance moved by the body after $2 s =$ Area $OAB ^{\prime} C ^{\prime}=2 m s ^{-1} \times 2 s =4 m$
Distance covered by the body after $12 s$
$=$ Area $OAED$ $+$ Area $BEF$ $+$ Area of $DHGI$ $+$ Area of $FHG$
$=2 m s ^{-1} \times 10 s +\frac{1}{2} \times 5 s \times 8 m s ^{-1}+$ $6 m s ^{-1} \times 2 s +\frac{1}{2} \times 2 s \times 5 m s ^{-1}$
$=20 m +20 m +12 m +4 m =56 m$
