Physics, asked by mehedihassan8208, 3 months ago

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

Answered by wwwkanikachatterjee
0

s = ut \:  + (1 \div 2)a {t }^{2} \\ u = 0 \\ s1 = 4 \times  {10 }^{2}  \\= 400 \\ v = at \\ 4 \times 10 = 40 \\ \\ s2 = vt \\  = 40 \times 10  = 400 \\  \\ v =  \sqrt{2as3}  \\ s3 =  {v}^{2}  \div 2g \\ 400 \times 400  \div 2 \times 10 \\ s3 = 8000 \\ s = s1 + s2 + s3 \\ s = 400 + 400 + 8000 \\ 8800 \\

hope it will help....

Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
3

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

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