Q28 / 46 · NTA Abhyas (JEE Main) 2020 · Physics · Motion In One Dimension

The figure shows the v-t graph of a particle moving in a straight line. Starting from rest, the time after which the particle returns to its starting position is

​Figure

  • Correct answer
    A
    36.2 s
     
  • B
    40 s
  • C
    25 s
  • D
    42.6 s
     

Solution

For a particle to return to origin. Net displacement is zero.
 
Figure

For t=0 to t=20 s,

a1=2020 =1 m s-2

For t=20 s to t s,

a2=0-2025-20=-4 m s-2

  0 = 1 2 × 2 0 × 2 5 + 1 2 × - 4 × t - 2 5 2

(t - 25)2 = 125
t - 25 = ± 11.2

t >25 s

t = 36.2 s

Hence, the correct answer is option: 'A'.

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