Question

Scooter moving at a speed of 10 meter per second is stopped by applying brake which produces a retardation of 5.5 meter per second square how much distance will be covered by the scooter before stop

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Initial Velocity = 10 m/s

✭ Final Velocity = 0 m/s

✭ Acceleration = 5.5 m/s²

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ Distance travelled?

$\displaystyle\large\underline{\sf\gray{Solution}}$

So here we may find the distance travelled with the help of the third Equation of motion, that is,

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

Where,

• v = Final Velocity = 0 m/s

• u = Initial Velocity = 10 m/s

• a = Acceleration = 5.5 m/s²

• s = Distance = ?

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

➳ $\displaystyle\sf v^2-u^2 = 2as$

➳ $\displaystyle\sf 0^2-10^2 = 2\times 5.5\times s$

➳ $\displaystyle\sf 0-100 = 11s$

➳ $\displaystyle\sf -100 = 11s$

➳ $\displaystyle\sf \dfrac{-100}{11} = s$

➳ $\displaystyle\sf Distance = -9.09 \ m$

➳ $\displaystyle\sf \pink{Distance = 9.09 \ m}$

Note :- Here we took the distance as 9.09m ignoring negative sign as distance can't be Negative (acceleration was). And also we considered the final velocity as 0 because the body comes to rest

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☯ $\displaystyle\underline{\large\sf Know \ More}$

⪼ $\displaystyle\sf v = u+at$

⪼ $\displaystyle\sf s = ut+\dfrac{1}{2} at^2$

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Answer 2

Answer:

Step (I) : Given data

• Velocity of the scooter ( Initial Velocity, u) = 10 m/s

• Acceleration = 5.5 m/s²

Step (II) : We've to Calculate the Distance traveled before the scotter stops

⇢ v² - u² = 2as

⇢ 0² - 10² = 2(5.5)s

⇢ -100 = 11s

⇢ s = -100/11

⇢ s = -9.09 m

As we know that Distance can't be negative so ignition negative sign distance travelled by scotter before it stops is 9.09 meter.