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A particle located  at x = 0 at time t = 0, starts moving along with the positive x-direction with a velocity 'v' that varies  as v = a√ x . The displacement of the particle varies with time as
  • a)
    t 2
  • b)
    t
  • c)
    t1/2
  • d)
    t 3
Correct answer is option 'A'. Can you explain this answer?
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A particle located at x = 0 at time t = 0, starts moving along with th...
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A particle located at x = 0 at time t = 0, starts moving along with th...
Explanation:

Given:
- Initial position, x = 0 at time t = 0
- Particle moves in the positive x-direction
- Velocity, v = a√x

Velocity-time relation:
- v = dx/dt = a√x
- dx = a√x dt
- Integrating both sides, ∫(1/√x) dx = ∫a dt
- 2√x = at + C
- At t = 0, x = 0, C = 0
- 2√x = at
- x = (a^2 t^2)/4

Displacement-time relation:
- x = (a^2 t^2)/4
- x = (a^2 t^2)/4 = a^2 (t^2)/4
- x ∝ t^2
Therefore, the displacement of the particle varies with time as t^2. Hence, the correct answer is option 'A'.
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