Question
The displacement of an object changes with time according to $s=t^3-6 t^2+3 t+4$, then the acceleration of the object at instant is zero. What is its velocity?
✓
Answer
Given :
$s=t^3-6 t^2+3 t+4$
$\begin{aligned}
\therefore \quad \text { Velocity } v & =\frac{ds}{dt}=\frac{d}{dt}\left(t^3-6 t^2+3 t+4\right) \\
v & =3 t^2-6 \times 2 t+3 \times 1+0 \\
v & =3 t^2-12 t+3 .......(1)\\
\text { Acceleration } a & =\frac{dv}{dt}=\frac{d}{dt}\left(3 t^2-12 t+3\right) \\
a & =3 \times 2 t-12 \times 1+0 \\
a & =6 t-12 ....(2)
\end{aligned}$
When acceleration $a =0$ then from equation (2)
$\begin{aligned}
6 t-12 & =0 \\
t & =2 \text { second }
\end{aligned}$
$t =2$ velocity per second from equ. (1)
$\begin{array}{l}
v=3(2)^2-12(2)+3 \\
v=12-24+3=-9 m / s
\end{array}$
Here the negative sign indicates that the displacement of the object is decreasing with time.
$s=t^3-6 t^2+3 t+4$
$\begin{aligned}
\therefore \quad \text { Velocity } v & =\frac{ds}{dt}=\frac{d}{dt}\left(t^3-6 t^2+3 t+4\right) \\
v & =3 t^2-6 \times 2 t+3 \times 1+0 \\
v & =3 t^2-12 t+3 .......(1)\\
\text { Acceleration } a & =\frac{dv}{dt}=\frac{d}{dt}\left(3 t^2-12 t+3\right) \\
a & =3 \times 2 t-12 \times 1+0 \\
a & =6 t-12 ....(2)
\end{aligned}$
When acceleration $a =0$ then from equation (2)
$\begin{aligned}
6 t-12 & =0 \\
t & =2 \text { second }
\end{aligned}$
$t =2$ velocity per second from equ. (1)
$\begin{array}{l}
v=3(2)^2-12(2)+3 \\
v=12-24+3=-9 m / s
\end{array}$
Here the negative sign indicates that the displacement of the object is decreasing with time.
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