1. Find the shortest distance between the pair of parallel
lines whose equations are
i = (i + 2j + 3 k) + lembda (i +3j + 4k)
and i = (2ỉ + 4ị + 5k) + meu (4ỉ + 6ị + 8k).
Answers
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.____________________________