Question

plz help this one also fast??? ​

Answer 1

$\sf\large\green{\underbrace{Answer\implies 236m}}}$

$\sf\large\underline\purple{Given:-}$

$\sf{\implies Intial\:_{(velocity)}=0}$

$\sf{\implies Final\:_{(velocity)}=24m/s}$

$\sf{\implies Time\:_{(take)}=19.6}$

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies Distance\:_{(traveled\:by\:car)}=?}$

$\sf\large\underline\purple{Solution:-}$

To calculate the distance traveled by car at first we have to represent the given value in abbreviation form so, intial velocity as you know that it the motion of any body in rest because of at a constant rate. Intial velocity=U , Final velocity=24m/s Time=t. After that by applying formula to calculate the acceleration of car then calculate distance traveled by car:-]

$\sf\large\underline\purple{Formula\:used:-}$

$\sf{\implies As\:per\: first\: equation\:of\: motion:-}$

$\tt{\implies v=u+at}$

$\tt{\implies 24=0+a*19.6}$

$\tt{\implies 19.6a=24-0}$

$\tt{\implies a=\dfrac{24}{19.6}}$

$\tt\red{\implies a=1.22\:m/s^2(approx)}$

$\sf{\implies As\:per\:3rd\: equation\:of\: motion:-}$

$\tt{\implies v^2-u^2=2as}$

$\tt{\implies 24^2-0^2=2(1.22)(s)}$

$\tt{\implies 576-0=2.44s}$

$\tt{\implies 2.44s=576}$

$\tt{\implies s=\dfrac{576}{2.44}}$

$\tt\red{\implies s=236\:m(approx)}$

$\sf\large{Hence,}$

$\sf{\implies Distance\:_{(traveled\:by\:car)}=236\:m}$

$\sf\green{\implies Some\: abbreviation\:form}$

$\sf{\implies A\:_{(stand\:for)}=acceleration:-}$

$\sf{\implies U\:_{(stand\:for)}=Intial\: velocity:-}$

$\sf{\implies V\:_{(stand\:for)}=Final\:velocity:-}$

$\sf{\implies S\:_{(stand\:for)}=Distance\:_{(travel\:by\:a\: body)}}$

Answer 2

Answer:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\sf\large\underline\blue{Given:-}$

$\sf{\implies Intial\:_{(velocity)}=0}$

$\sf{\implies Final\:_{(velocity)}=24m/s}$

$\sf{\implies Time\:_{(take)}=19.6}$

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies Distance\:_{(traveled\:by\:car)}=?}$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\large\underline\blue{Formula\:used:-}$

$\sf{\implies As\:per\: first\: equation\:of\: motion:-}$

$\tt{\implies v=u+at}$

$\tt{\implies 24=0+a*19.6}$

$\tt{\implies 19.6a=24-0}$

$\tt{\implies a=\dfrac{24}{19.6}}$

$\tt\red{\implies a=1.22\:m/s^2(approx)}$

$\sf{\implies As\:per\:3rd\: equation\:of\: motion:-}$

$\tt{\implies v^2-u^2=2as}$

$\tt{\implies 24^2-0^2=2(1.22)(s)}$

$\tt{\implies 576-0=2.44s}$

$\tt{\implies 2.44s=576}$

$\tt{\implies s=\dfrac{576}{2.44}}$

$\tt\orange{\implies s=236\:m(approx)}$

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$\sf\large{Hence,}$

$\sf{\implies Distance\:_{(traveled\:by\:car)}=236\:m}$

$\sf\pink{\implies Some\: abbreviation\:form}$

$\sf{\implies A\:_{(stand\:for)}=acceleration:-}$

$\sf{\implies U\:_{(stand\:for)}=Intial\: velocity:-}$

$\sf{\implies V\:_{(stand\:for)}=Final\:velocity:-}$

$\sf{\implies S\:_{(stand\:for)}=Distance\:_{(travel\:by\:a\: body)}}$

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