Question

A scooter moving at a certain initial velocity comes to rest in 62.5 m.If the retardation due to applied brakes was 5m/s².Find the initial velocity of the scooter
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Answer 1

➤ Answer

• Initial velocity of the scooter = 25m/s.

➤ Given

• Final velocity = 0 m/s.
• Distance = 62.5 meters.
• Acceleration ( retardation ) = -5m/s².

➤ To Find

• The initial velocity of the scooter.

➤ Step By Step Explanation

Here we need to find the initial velocity of the scooter. So let's do it !!

➠ By using third equation of motion

• v² - u² = 2as

➠ By substituting the values

$\longmapsto\tt {v}^{2} - {u}^{2} = 2as \\ \\ \longmapsto\tt {(0)}^{2} - {u}^{2} = 2 \times - 5 \times 62.5 \\ \\ \longmapsto\tt - {u}^{2} = - 625 \\ \\\longmapsto\tt {u}^{2} = 625 \\ \\\longmapsto\tt u = \sqrt{625} \\ \\\longmapsto\bf{\green{ u = 25\:m/s}}$

➵ Therefore, initial velocity of the scooter = 25m/s.

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➤ Equations of Motion :

There are mainly three equations of motion. They are as follows -

• v = u + at
• s = ut + ½at²
• v² - u² = 2as

Where, u = initial velocity , v = final velocity, t = time, a = acceleration, and s = distance travelled.

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Answer 2

$\underbrace{\underline{\underline{\sf{\maltese\:Given\:Information:-}}}}$

$\qquad\tt{:\implies\:Final\:velocity\:=\:0\:m/s}$

$\qquad\tt{:\implies\:Distance\:=\:62.5\:m}$

$\qquad\tt{:\implies\:Retardation\:=\:-\:5\:m/s^{2}}$

$\underbrace{\underline{\underline{\sf{\maltese\:To\:Find\:Out:-}}}}$

$\qquad\tt{:\implies\:Initial\:velocity\:of\:the\:body\:.}$

$\underbrace{\underline{\underline{\sf{\maltese\:Using\:Formula:-}}}}$

$\qquad\tt{:\implies\:v^{2}\:-\:u^{2}\:=\:2as}$

$\underbrace{\underline{\underline{\sf{\maltese\:Required\:Solution:-}}}}$

$\qquad\bf{:\implies\:v^{2}\:-\:u^{2}\:=\:2as}$

$\qquad\bf{:\implies\:(\:0\:)^{2}\:-\:u^{2}\:=\:2\:\times\:-5\:\times\:62.5}$

$\qquad\bf{:\implies\:-\:u^{2}\:=\:-\:625}$

$\qquad\bf{:\implies\:u^{2}\:=\:625}$

$\qquad\bf{:\implies\:u\:=\:\sqrt{625}}$

$\qquad\bf{:\implies\:u\:=\:25\:m/s}$

Hence the initial velocity of the scooter is 25m/s.