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[tex]\sf\bold{The\:above\:graph\:says:}[/tex]
- $\sf\bold\red{a=slope\:of\:v-t\:graph}$
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$\sf\bold{here:}$
- $\sf\bold\red{a=acceleration}$
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$\sf\bold{Find\:acceleration\:by\:v-t\:graph:}$
$\mapsto$ $\sf\small{TanØ =}$ $\sf\dfrac{slope\:of\:y}{slope\:of\:x}$= $\sf\dfrac{20}{8}$ $\sf{m/s^2}$
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- $\mapsto$ [tex]\sf\underline\bold\purple{a=2.5m/s^2}[/tex]
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$\sf\bold{Now,calculate\:the\:distance:}$
- $\sf\bold{By\:equation\:of\:motion.}$
- $\sf{s=}$ $\sf{ut+}$ $\sf\dfrac{1}{2}$ $\sf{at^2}$
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$\sf\bold{Given\:parameters:}$
- $\sf\bold{u=0}$
- $\sf\bold{a=5/8}$
- $\sf\bold{t=8}$
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$\sf\bold{Substituting \: the\:values:}$
$\mapsto$ $\sf\small{s=0+}$ $\sf\dfrac{1}{2}$ $\times$ $\sf\dfrac{5}{8}$ $\sf\small{8^2}$
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$\longmapsto$ $\sf\underline\bold\purple{S=80m}$
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❍Therefore , the acceleration is $\sf\bold{2.5m/s^2}$ and the distance traveled by car is $\sf\bold{80m.}$
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