Question

An object experiences an acceleration of -6.8 m/s2. As a result, it accelerates from 54 m/s to a complete stop. How much distance did it travel during that acceleration? SHOW ALL WORK!

Answer 1

$\huge\sf\pink{Answer}$

☞ Your Answer is 224.3 m

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$\huge\sf\blue{Given}$

✭ Acceleration = -6.8 m/s²

✭ Initial Velocity = 54 m/s

✭ Final Velocity = 0 m/s

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$\huge\sf\gray{To \:Find}$

◈ The distance travelled?

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$\huge\sf\purple{Steps}$

$\sf \large\underline{\underline{\sf Acceleration}}$

»» Acceleration is a vector quantity,i.e it has both magnitude and direction

»» Acceleration is equal to the change in the velocity of a body

»» A = v-u/t

»» SI unit of acceleration is m/s²

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So here we may use the third Equation of motion, i.e,

$\underline{\boxed{\sf v^2-u^2 = 2as}}$

◕ v = Final Velocity

◕ u = Initial Velocity

◕ a = Acceleration

◕ s = Distance travelled

Substituting the given values,

➝ $\sf v^2-u^2 = 2as$

➝ $\sf 0^2-54^2 = 2\times (-6.8) \times s$

➝ $\sf 0 = 2\times (-6.8) \times s + 54^2$

➝ $\sf 0 = -13.6 \times s+2916$

➝ $\sf -2916 = -13.6s$

➝ $\sf \dfrac{-2916}{-13.6} = s$

➝ $\sf \orange{s = 224.3 \ m}$

☯ $\sf \underline{\sf Know \ More}$

⪼ v = u+at

⪼ s = ut+½at²

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