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Physics · MOTION IN A STRAIGHT LINE
हिन्दीThe speed of a particle moving along a straight line varies with time as \( v(t) = (4t - t^2)\text{ m s}^{-1} \) for \( 0 \le t \le 4\text{ s} \). What are the magnitude of average acceleration during the time interval \( t = 0 \) to \( t = 3\text{ s} \) and the magnitude of instantaneous acceleration at \( t = 3\text{ s} \)?
Correct answer · Option 1
\( 1\text{ m s}^{-2} \) and \( 2\text{ m s}^{-2} \)
Explanation
Average acceleration = \( \frac{v(3)-v(0)}{3-0} = \frac{3-0}{3} = 1\text{ m s}^{-2} \). Instantaneous acceleration \( a(t) = \frac{dv}{dt} = 4 - 2t \). At \( t=3 \), \( a(3) = 4 - 6 = -2\text{ m s}^{-2} \), so its magnitude is \( 2\text{ m s}^{-2} \).
