A particle starts form rest on a straight line. First it accelerates at 4 m/s2 for 10 seconds. Then for 10 seconds it moves with the acquired constant velocity. Then it decelerates at 8 m/s2 to rest. Total distance covered during the entire journey is
Answers
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Given
- Initial Velocity = 0 m/s
- Acceleration₁ = 4 m/s
- Time= 10 sec
- Later on it travels with a constant acceleration for 10 sec
- Then it starts regarding at a rate of 8 m/s²
To Find
- Distance Covered before it goes to rest
Solution
☯ v = u + at [First Equation of Motion]
☯ v²-u² = 2as [Third Equation of Motion]
☯ s = ut + ½at² [Second Equation of Motion]
✭ Final Velocity after the acceleration :
→ v = u + at
→ v = 0 + 4 × 10
→ v = 0 + 40
→ v = 40 m/s
✭ Distance Covered during Accn :
→ v²-u² = 2as
→ 40²-0² = 2 × 4 × s
→ 1600 = 4s
→ 1600/4 = s
→ Distance (Accn) = 400 m
✭ Distance Covered during constant Accn :
→ s = ut + ½at²
→ s = 40×10 + ½ × 0 × 10²
→ s = 400 + 0
→ Distance (Constant Velocity) = 400 m
✭ Distance Covered during retardation :
→ v²-u² = 2as
→ 0²-40² = 2 × (-8) × s
→ -1600 = -16s
→ -1600/-16 = s
→ Distance (Retardation) = 100 m
✭ Total Distance :
→ Total Distance = Distance during acceleration + Distance during Constant Velocity + Distance during deceleration
→ Total Distance = 400 + 400 + 100
→ Total Distance = 900 m