The $v-t$ graph of cars $A$ and $B$ which start from the same place and move along a straight road in the same direction is shown. Calculate:
$(i)$ The acceleration of car $A$ between $0$ and $8\, s$.
$(ii)$ The acceleration of car $B$ between $2\, s$ and $4\, s$.
$(iii)$ The points of time at which both the cars have the same velocity.
$(iv)$ Which of the two cars is ahead after $8\, s$ and by how much?

(N/A) $(i)$ Acceleration of car $A$ between $0$ and $8\, s$ is the slope of the line:
$a = \frac{v_f - v_i}{t_f - t_i} = \frac{80 - 0}{8 - 0} = 10\, m/s^2$.
$(ii)$ Acceleration of car $B$ between $2\, s$ and $4\, s$ is the slope of the line:
$a = \frac{v_f - v_i}{t_f - t_i} = \frac{60 - 20}{4 - 2} = \frac{40}{2} = 20\, m/s^2$.
$(iii)$ Both cars have the same velocity where the two lines intersect,which occurs at $t = 2\, s$ and $t = 6\, s$.
$(iv)$ Distance travelled is the area under the $v-t$ graph.
Distance travelled by car $A$ in $8\, s$ = Area of triangle with base $8\, s$ and height $80\, m/s$ = $\frac{1}{2} \times 8 \times 80 = 320\, m$.
Distance travelled by car $B$ in $8\, s$ = Area of triangle ($t=1$ to $2$) + Area of trapezium ($t=2$ to $4$) + Area of rectangle ($t=4$ to $8$).
Area = $(\frac{1}{2} \times 1 \times 20) + (\frac{20+60}{2} \times 2) + (4 \times 60) = 10 + 80 + 240 = 330\, m$.
Since $330\, m > 320\, m$,car $B$ is ahead by $330 - 320 = 10\, m$.