The velocity-time graph of cars $A$ and $B$ which start from the same place and move along a straight road in the same direction is shown below

Calculate :

$(a)$ the acceleration of car $B$ between $2 \,s$ and $4\, s$.

$(b)$ the time at which both the cars have the same velocity.

$(c)$ the distance travelled by the two cars $A$ and $B$ in $8\, s$

$(d)$ Which of the two cars is ahead after $8\, s$ and by how much ?

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$(a)$ $a=$ slope of $v-t$ graph $=\frac{40-20}{4-2}=\frac{20}{2}$

$=10 m s ^{-2}$

$(b)$ $2$ second

$(c)$ Car $A : 390 m ,$ Car $B : 320 m$

Distance $=$ Area under $v-t$ graph

Car $B$ $x=\frac{1}{2} \times 8 \times 80=320 m$

Car $A$ $x=\left(\frac{1}{2} \times 3 \times 60\right)+(60 \times 5)$

$=90+300$

$=390 m$

$(d)$ Car $B,$ $20 m$

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