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Motion in A Straight Line question

2023 · 6 Apr · Shift 2 · Q45
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Motion in A Straight Line question

2023 · 6 Apr · Shift 2 · Q45

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
A particle starts with an initial velocity of 10.0 ms−110.0 \mathrm{~ms}^{-1}10.0 ms−1 along xxx-direction and accelerates uniformly at the rate of 2.0 ms−22.0 \mathrm{~ms}^{-2}2.0 ms−2. The time taken by the particle to reach the velocity of 60.0 ms−160.0 \mathrm{~ms}^{-1}60.0 ms−1 is ‾\underline{\hspace{2cm}}​.
  1. A
    30s
  2. B
    6s
  3. C
    3s
  4. D
    25s
View written solutionFree

Correct answer: D

  1. Given data

    • Initial velocity: u=10 m s−1u = 10\,\text{m s}^{-1}u=10m s−1
    • Final velocity: v=60 m s−1v = 60\,\text{m s}^{-1}v=60m s−1
    • Uniform acceleration: a=2 m s−2a = 2\,\text{m s}^{-2}a=2m s−2

  2. Use the first equation of motion v=u+atv = u + atv=u+at

  3. Substitute the values 60=10+2t60 = 10 + 2t60=10+2t

  4. Solve for ttt 60−10=2t60 - 10 = 2t60−10=2t 50=2t50 = 2t50=2t t=25 st = 25\,\text{s}t=25s

  5. Match with the options 25 s25\,\text{s}25s corresponds to Option D.

  6. Comparison with stored answer Stored correct answer: D

    My derived answer: D

    Hence, the derived answer agrees with the stored answer.

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