An object is moving along a straight line with uniform acceleration. The following table gives the velocity of the object at various instants of time.

Time $(s)$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
Velocity $(m s^{-1})$	$2$	$4$	$6$	$8$	$10$	$12$	$14$

Plot the graph.
From the graph:
$(i)$ Find the velocity of the object at the end of $2.5 \ s$.
$(ii)$ Calculate the acceleration.
$(iii)$ Calculate the distance covered in the last $4 \ s$.

(N/A) $(i)$ By observing the velocity-time graph, at $t = 2.5 \ s$, the velocity is $7 \ m s^{-1}$.
$(ii)$ Acceleration $(a)$ is the slope of the velocity-time graph.
$a = \frac{v - u}{t} = \frac{14 - 2}{6 - 0} = \frac{12}{6} = 2 \ m s^{-2}$.
$(iii)$ The distance covered in the last $4 \ s$ (from $t = 2 \ s$ to $t = 6 \ s$) is equal to the area under the velocity-time graph between these time intervals.
This area forms a trapezium $ABCD$ where the parallel sides are the velocities at $t = 2 \ s$ $(6 \ m s^{-1})$ and $t = 6 \ s$ $(14 \ m s^{-1})$, and the height is the time interval $(6 - 2 = 4 \ s)$.
$S = \text{Area of trapezium } ABCD = \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
$S = \frac{1}{2} \times (6 + 14) \times 4 = \frac{1}{2} \times 20 \times 4 = 40 \ m$.