Question

A scooter moving at a speed of 10m/s is
stopped by applying brakes which produce
acceleration of -0.5m/s. How much distance
will be covered by the scooter?


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Answer 1

Answer:

A scooter moving at a speed of 10m/s is

stopped by applying brakes which produce

acceleration of -0.5m/s. How much distance

will be covered by the scooter?

Answer 2

$\large\underline{\red{\sf \purple{\bigstar} .}}$Given :-

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• A scooter moving at a speed of 10m/s is stopped by applying brakes which produce acceleration of -0.5m/s².

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$\large\underline{\red{\sf \blue{\bigstar} .}}$To find :-

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• How much distance will be covered by the scooter?

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$\large\underline{\red{\sf \red{\bigstar} .}}$Solution :-

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• Initial velocity = 10 m/s

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• Acceleration = - 0.5 m/s²

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• Final velocity = 0

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According to third equation of motion

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$\large\underline{\red{\sf \green{\bigstar} .}}$→ v² = u² + 2as

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• Where " v " is final velocity, " u " is initial velocity, " a " is acceleration and " s " is distance.

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• → (0)² = (10)² + 2 × (-0.5) × s

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• → 0 = 100 - 1 × s

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• → 0 = 100 - s

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$\large\underline{\green{\sf \pink{\bigstar} .}}$→ s = 100 m

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$\large\underline{\red{\sf \orange{\bigstar} .}}$Hence,

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$\large\underline{\red{\sf \red{\bigstar} .}}$Distance will be covered by the Scooter is 100m

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$\large\underline{\red{\sf \purple{\bigstar} .}}$More to know $\large\underline{\red{\sf \green{\bigstar} .}}$

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• s = ut + ½ at² (second equation of motion)

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• v = u + at (first equation of motion)

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• Acceleration is the rate of change in velocity.

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• Acceleration in negative is known as retardation.

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