Answer
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Hint: Here displacement is given from that we can find velocity and acceleration in terms of time t. As differentiating displacement with respect to time we get velocity and differentiating velocity with respect to time we get acceleration. And it is given that we have to find velocity when acceleration is zero.
Formula used:
\[\begin{align}
& v=\dfrac{ds}{dt} \\
& a=\dfrac{dv}{dt} \\
\end{align}\]
Complete step-by-step solution:
We know that velocity is the rate of change of displacement with respect of time or it is given as displacement by time. Here according to question we will use the rate of change of displacement with respect to time.
\[\begin{align}
& v=\dfrac{ds}{dt} \\
& v=\dfrac{d({{t}^{3}}-6{{t}^{2}}+3t+4)}{dt} \\
& v=3{{t}^{2}}-12t+3 \\
\end{align}\]
Now the acceleration is given as rate of change of velocity with respect to time.
\[\begin{align}
& a=\dfrac{dv}{dt} \\
& a=\dfrac{d(3{{t}^{2}}-12t+3)}{dt} \\
& a=6t-12 \\
\end{align}\]
We have to find velocity when acceleration is zero. Hence equating the above equation to zero.
\[\begin{align}
& 6t-12=0 \\
& \Rightarrow 6t=12 \\
& \Rightarrow t=2 \\
\end{align}\]
Substituting the value of t=2 in the velocity equation we get
\[\begin{align}
& \Rightarrow v=3{{(2)}^{2}}-12(2)+3 \\
& \Rightarrow v=3(4)-24+3 \\
& \Rightarrow v=12-24+3 \\
& \Rightarrow v=-9m/s \\
\end{align}\]
Hence when acceleration is equal to zero the velocity will be -9m/s. Therefore the correct option is D.
Note: The negative value of velocity shows the decrease in speed. Acceleration can also have a negative sign which shows retardation that is the velocity decreases and the object is slowing down.
While differentiating power should be changed properly neither the answer will be wrong. We can also observe that acceleration is a double differentiation of displacement.
Formula used:
\[\begin{align}
& v=\dfrac{ds}{dt} \\
& a=\dfrac{dv}{dt} \\
\end{align}\]
Complete step-by-step solution:
We know that velocity is the rate of change of displacement with respect of time or it is given as displacement by time. Here according to question we will use the rate of change of displacement with respect to time.
\[\begin{align}
& v=\dfrac{ds}{dt} \\
& v=\dfrac{d({{t}^{3}}-6{{t}^{2}}+3t+4)}{dt} \\
& v=3{{t}^{2}}-12t+3 \\
\end{align}\]
Now the acceleration is given as rate of change of velocity with respect to time.
\[\begin{align}
& a=\dfrac{dv}{dt} \\
& a=\dfrac{d(3{{t}^{2}}-12t+3)}{dt} \\
& a=6t-12 \\
\end{align}\]
We have to find velocity when acceleration is zero. Hence equating the above equation to zero.
\[\begin{align}
& 6t-12=0 \\
& \Rightarrow 6t=12 \\
& \Rightarrow t=2 \\
\end{align}\]
Substituting the value of t=2 in the velocity equation we get
\[\begin{align}
& \Rightarrow v=3{{(2)}^{2}}-12(2)+3 \\
& \Rightarrow v=3(4)-24+3 \\
& \Rightarrow v=12-24+3 \\
& \Rightarrow v=-9m/s \\
\end{align}\]
Hence when acceleration is equal to zero the velocity will be -9m/s. Therefore the correct option is D.
Note: The negative value of velocity shows the decrease in speed. Acceleration can also have a negative sign which shows retardation that is the velocity decreases and the object is slowing down.
While differentiating power should be changed properly neither the answer will be wrong. We can also observe that acceleration is a double differentiation of displacement.
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