A particle moves along a straight line. At t = 0, it is at rest. Its acceleration varies with time according to a(t) = 6t - 2, where a is in m/ $s^2$ and t is in seconds. Considering the motion for t >= 0, the minimum value of the velocity attained by the particle is:
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Question

A particle moves along a straight line. At t = 0, it is at rest. Its acceleration varies with time according to a(t) = 6t - 2, where a is in m/ $s^2$ and t is in seconds. Considering the motion for t >= 0, the minimum value of the velocity attained by the particle is:

Options

G1 Physics · Accepted Answer

Since a = dv/dt = 6t - 2, integrating gives v = 3t^2 - 2t + C. As v(0) = 0, C = 0. Velocity is minimum when dv/dt = a = 0, so 6t - 2 = 0 and t = 1/3 s. Therefore v_min = 3(1/9) - 2(1/3) = 1/3 - 2/3 = -1/3 m/s. Hence, option A is correct.

A particle moves along a straight line. At t = 0, it is at rest. Its acceleration varies with time according to a(t) = 6t - 2, where a is in m/s^2 and t is in seconds. Considering the motion for t >= 0, the minimum value of the velocity attained by the particle is: