Question

A scooter moving at a speed of 10m/s is stopped by applying brakes which produce acceleration of -0.5m/s². How much distance will be covered by the scooter?​

Answer 1

Given:-

• Initial velocity of scooter (u) =10m/s

• Acceleration of the scooter (a) = -0.5m/s²

• Final velocity of scooter (v) = 0m/s

To Find:-

• Distance covered by scooter (s).

Solution:-

By using third equation of motion

⟹ v² = u² +2as

Put all the values which is given above

⟹ 0² = 10² + 2 × (-0.5) ×s

⟹ 0 = 100 + (-1 ) ×s

⟹ -100 = -1 ×s

⟹ s = -100/-1

⟹ s = 100m

∴ The Distance covered by the scooter is 100m.

Additional Information!!

Some equation of motion are

⟹ v = u +at

⟹s = ut +1/2at².

⟹ v² = u² + 2as

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement or Distance

t is the time taken

Answer 2

Aɴꜱᴡᴇʀ :

➛ The distance covered by the scooter = 100m.

Gɪᴠᴇɴ :

• Initial velocity (u) = 10 m/s
• Acceleration (a) = -0.5 m/s²
• Final velocity (v) = 0 m/s

Tᴏ Fɪɴᴅ :

• The distance covered by the scooter = ?

Sᴏʟᴜᴛɪᴏɴ :

From the third equation of motion

➝ v² = u² + 2as

➝ 0² = (10)² + 2 × (-0.5) × s

➝ 0 = 100 + (-1) × s

➝ s = -100/-1

➝ s = 100 m

Hence, the distance covered by the scooter is 100m.

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Some releted formula

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as $\:$

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