$A$ body moves with a velocity of $2\, m s^{-1}$ for $5\, s$,then its velocity increases uniformly to $10\, m s^{-1}$ in the 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 $12\, s$.

(N/A) $(ii)$ According to the velocity-time graph:
Distance moved by the body after $2\, s$ is equal to the area under the graph from $t = 0$ to $t = 2\, s$.
Area $= 2\, m s^{-1} \times 2\, s = 4\, m$.
Distance covered by the body after $12\, s$ is the total area under the graph from $t = 0$ to $t = 12\, s$.
This area consists of the rectangle for the first $5\, s$,the trapezoid for the next $5\, s$,and the trapezoid for the final $2\, s$ (from $t = 10\, s$ to $t = 12\, s$).
Total Distance $= (2\, m s^{-1} \times 5\, s) + \frac{1}{2} \times (2\, m s^{-1} + 10\, m s^{-1}) \times 5\, s + \frac{1}{2} \times (10\, m s^{-1} + 6\, m s^{-1}) \times 2\, s$
$= 10\, m + 30\, m + 16\, m = 56\, m$.