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Physics · MOTION IN A STRAIGHT LINE
हिन्दीThe position of a particle moving along a straight line is given by \( x(t) = \alpha t^3 - \beta t^2 + \gamma t \), where \( \alpha = 0.08 \text{ m/s}^3 \), \( \beta = 0.6 \text{ m/s}^2 \), and \( \gamma = 3.6 \text{ m/s} \). Determine the time \( t \) at which the instantaneous acceleration of the particle becomes zero, and its instantaneous velocity at that moment.
Correct answer · Option 4
\( t = 2.5 \text{ s}, v = 2.10 \text{ m/s} \)
Explanation
Velocity \( v(t) = \frac{dx}{dt} = 3\alpha t^2 - 2\beta t + \gamma = 0.24 t^2 - 1.2 t + 3.6 \). Acceleration \( a(t) = \frac{dv}{dt} = 0.48 t - 1.2 \). Setting \( a(t) = 0 \) gives \( t = 2.5 \text{ s} \). Substituting \( t = 2.5 \text{ s} \) into \( v(t) \) yields \( v = 0.24(2.5)^2 - 1.2(2.5) + 3.6 = 2.10 \text{ m/s} \).
