Question

Elaborate or expound the statement in a paragraph form with a
maximum of 5 sentences
I have learned that when researching online, I have to remember to​

Answer 1

before stopping (v=0)

\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}

∣ANSWER−∣

Distance travelled before stopping is 12m

\Large{\underline{\underline{\bf{Given:-}}}}

Given:−

Initial velocity = 43.2km/h

Initial velocity = 43.2km/hAcceleration = - 6m/s²

\Large{\underline{\underline{\bf{Find:-}}}}

Find:−

Distance travelled before stopping ?

Distance travelled before stopping ?Here we go yo ^^":−

[tex]\Large{\underline{\underline{\bf{SoLuTion:-}}}} [tex]\Large{\underline{\underline{\bf{SoLuTion:-}}}}[tex]

SoLuTion:−

[tex]

SoLuTion:−

❥ Final velocity = 43.2 km/h

❥ Final velocity = 43.2 km/h❥ v = 12 m/s

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity" a " = acceleration

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity" a " = acceleration" s " = distance

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity" a " = acceleration" s " = distance❥ (0)² = (12)² + 2 × (-6) × s

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity" a " = acceleration" s " = distance❥ (0)² = (12)² + 2 × (-6) × s❥ 0 = 144 - 12s

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity" a " = acceleration" s " = distance❥ (0)² = (12)² + 2 × (-6) × s❥ 0 = 144 - 12s❥ 12s = 144

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity" a " = acceleration" s " = distance❥ (0)² = (12)² + 2 × (-6) × s❥ 0 = 144 - 12s❥ 12s = 144❥ s = 144/12

❥ Final velocity = 43.2 km/h❥ v = 12 m/sApplying Third law of motion↠ v² = u² + 2as" v " = final velocity" u " = initial velocity" a " = acceleration" s " = distance❥ (0)² = (12)² + 2 × (-6) × s❥ 0 = 144 - 12s❥ 12s = 144❥ s = 144/12❥ s = 12 m

\mathfrak{\huge{\purple{\underline{\underline{Hence}}}}}

Hence

S = 12m

\Large{\underline{\underline{\bf{Additional Information:-}}}}

AdditionalInformation:−

❥ In the first law, an object will not change its motion unless a force acts on it.

❥ In the first law, an object will not change its motion unless a force acts on it.❥ In the second law, the force on an object is equal to its mass times its acceleration.

❥ In the first law, an object will not change its motion unless a force acts on it.❥ In the second law, the force on an object is equal to its mass times its acceleration.❥ In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

❥ In the first law, an object will not change its motion unless a force acts on it.❥ In the second law, the force on an object is equal to its mass times its acceleration.❥ In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.____________________________