Solution:
The particle comes to rest when \(v = 0\), giving \(4t - 2 = 0 ⇒ t = 0.5\text{ s}\). Integrating velocity to find position, \(x = int_0^t (4t - 2)dt = 2t^2 - 2t\). At \(t = 0.5\text{ s}\), \(x = 2(0.5)^2 - 2(0.5) = -0.5\text{ m}\).
A particle moves in a straight line with velocity \(v = (4t - 2)\text{ m/s}\). If the particle starts from \(x = 0\), then the position of particle at which it momentarily comes to rest is
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