Thank you for reporting, we will resolve it shortly
Q.
A particle moves along a straight line such that its displacement at any time t is given by $ S={{t}^{3}}-6{{t}^{2}}+3t+4. $ The velocity when the accelerations zero in m/s is:
Given $ s={{t}^{3}}-6{{t}^{2}}+3t+4 $ $ \frac{ds}{dt}=3{{t}^{2}}-12t+3 $ and $ \frac{{{d}^{2}}s}{d{{t}^{2}}}=6t-12 $ If acceleration is zero, $ 6t-12=0 $ or $ t=2\sec $ Hence, velocity at t = 2 sec $ v=\frac{ds}{dt}=3{{t}^{2}}-12t+3 $ $ =12-24+3=-\,9\,m\text{/}s $