Question

A car accelerates (at a constant rate) from rest to 24-m/s in 19.6-seconds. How far will it travel during that time?
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Answer 1

$\huge\blue{\tt Given :-}$

• A car accelerates (at a constant rate) from rest to 24 m/s in 19.6 seconds

$\huge\orange{\tt To \: Find :-}$

• Distance traveled by car

$\huge\red{\tt Solution :-}$

We have,

• Initial Velocity(u) = 0 m/s
• Final Velocity(v) = 24 m/s
• Time taken(t) = 19.6 s

$\green{\bigstar} \: \underline{\tt Acceleration \: of \: car :-}$

We know that,

$\boxed{\tt v = u + at } \: \: [ \bf 1st \: equation \: of \: motion ]$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time taken

Substituting the given values we get,

$:\implies\tt 24 = 0 + a(19.6)$

$:\implies\tt 24 = 0 + 19.6a$

$:\implies\tt - 19.6a = 0 - 24$

$:\implies\tt\cancel- 19.6a = \cancel- 24$

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

$:\implies\green{\tt a = 1.22 \: m/s^{2}}$

$\green{\bigstar} \: \underline{\tt Distance \: traveled \: by \: car :-}$

We know that,

$\boxed{\tt v^{2} - u^{2} = 2as} \: \: [ \bf 3rd \: equation \: of \: motion ]$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance traveled

Substituting the values we get,

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

$:\implies\tt 576 = 2.44 \: s$

$:\implies\tt\cancel- 2.44 \: s = \cancel- 576$

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

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

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Additional Information :-

★ 2nd equation of motion :-

$\boxed{\tt s = ut + \dfrac{1}{2} at^{2}}$

Where,

• s = Distance Covered
• u = Initial Velocity
• t = Time taken

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Answer 2

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

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

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

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

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

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

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

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

• 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:-]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sf\large{Hence,}$

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

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

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

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

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

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