The $v-t$ graph of cars $A$ and $B$ which start from the same place and move along 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 ?

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$(i)$ Acceleration of car $A$ between $0$ to $8\, s$ is

$a=\frac{\text { speed }}{\text { time }}=\frac{80}{8}=10 m s ^{-2}$

$(ii)$ Acceleration of car $B$ between $2$ to $4\, s$ is

$a=\frac{\text { speed }}{\text { time }}=\frac{(60-20)}{2}=20 m s ^{-2}$

$(iii)$ After $2\, s$ and $6\, s$ from the start.

$(iv)$ Distance travelled by $CMA$

Area of $\Delta OAB =\frac{1}{2} \times 80 \times 8$

$=320 m$

Distance travelled by car $B$

Area of the trapezium $CDEB$

$=\frac{1}{2} \times(7+4) \times 60$

$=330 m$

Car $B$ is ahead by $10\, m$.

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