Question

An object undergoes an acceleration of 8m/sec² starting from rest find distance traveld in 1 sec​

Answer 1

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Acceleration\: (a) = 8 \:m/s^{2}}$

$\:\:\:\:\bullet\:\:\:\sf{Initial\: velocity \:(u) = 0 \:m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Time \:(t) = 1\: seconds}$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Distance \:travelled \:(s)}$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}}$

☯ Using 2nd equation of motion

$\dashrightarrow\:\: \sf{s = ut + \frac{1}{2}a {t}^{2} }$

$\dashrightarrow\:\: \sf{0 \times 1+ \frac{1}{ \cancel2} \times \cancel8 \times 1\times 1}$

$\dashrightarrow\:\: \sf{s = 4\times 1 \times 1}$

$\dashrightarrow\:\: {\boxed{\sf{s = 4\:m}}}$

${\mathfrak{\underline{\purple{\:\:\: Additional\: Information:-\:\:\:}}}}$

☣ Equations of motion

$\boxed{</p><p></p><p>\begin{minipage}{3 cm} \\</p><p></p><p>\sf{\:\bullet\:\:v = u +at} \\ \\</p><p></p><p>\sf{\:\bullet\:\:s = ut + \frac{1}{2}\:at^{2} }\\ \\</p><p></p><p>\sf{\:\bullet\:\:v^{2} = u^{2} + 2as}\\ \\</p><p>\sf{\:\bullet\:\:s = \dfrac{1}{2} (u + v)t}\\ </p><p></p><p>\end{minipage}</p><p></p><p>}$

$\sf{Where,}$

$\:\:\:\:\bullet\:\:\:\textsf{v = Final velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{u = Initial velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{a = Acceleration}$

$\:\:\:\:\bullet\:\:\:\textsf{s = Distance}$

$\:\:\:\:\bullet\:\:\:\textsf{t = Time taken}$

Answer 2

Answer:

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As per as the provided Question , It is asked to find the distance traveled by an object.

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$\large \bigstar\bf \underline \color{navy}{concept - \: } \color{black}\bigstar$

An object ( suppose a car ) makes an Acceleration of 8m/sec². It has started from rest , rest is always zero ( rest = 0 ) As Newton’s first law tells that, at rest is just a velocity vector of zero. Time taken is 1 second.

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$\large \bigstar\bf \underline \color{navy}{Given - \: } \color{black}\bigstar$

• Acceleration of an object , a = 8 m/sec²
• Initial velocity , u = 0
• Time taken , t = 1 sec

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$\large \bigstar\bf \underline \color{navy}{To find - \: } \color{black}\bigstar$

• Distance traveled , s = ?

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$\large \bigstar\bf \underline \color{navy}{Formula\: used - \: } \color{black}\bigstar$

$\large \dagger\bf \color{orange}{s = ut + \frac{1}{2} at {}^{2} } \color{black}\dagger$

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$\large \bigstar\bf \underline \color{navy}{Solution - \: } \color{black}\bigstar$

$\\ \sf \color{purple}{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > s = 0 \times 1 + \frac{1}{2} \times 8 \times (1) {}^{2} } \\ \sf \green { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > s = 0 + \frac{1}{ \cancel2} \times \cancel 8 {}^{4} \times 1} \\ \sf \red{ = > s = 4 \times 1} \\ \large \sf {\boxed {\boxed { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > s = 4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}}$

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$\\ \therefore\bf \color{blue}{the \: distance \: traveled \: by \: an \: object \: is \: 4m}$

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