(a) If a particle’s position is given by $\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{-}\mathbf{12}\mathbf{t}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}$(where tis in sec and x in metres), what is its velocity at $\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{sec}$? (b) Is it moving in positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculations) (e) Is there ever an instant when the velocity is zero? If so, give the time t, if no, answer no. (f) Is there a time after $\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{}\mathbf{sec}$when the particle is moving in negative direction of x? If so, give the time t, if no, answer no.

The position of the particle is given by the equation,

$\mathrm{x}=4-12\mathrm{t}+3{\mathrm{t}}^2$

The velocity can be found by differentiating the displacement with respect to time. Once the equation is derived for the velocity, substitute the values of time and displacement to find the value. The equation for the velocity in terms of displacement is,

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}$

Take the derivative of x to find velocity to get,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}} \\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(4-12\mathrm{t}+3{\mathrm{t}}^2\right) \\ =-12+6\left(\mathrm{t}\right) \end{gathered}$$

Therefore,

$\mathrm{v}=-12+6\left(\mathrm{t}\right)\left(\mathrm{i}\right)$

Substitute the value of t to find velocity.

$\mathrm{v}=-6\hspace{0.33em}\mathrm{m}/\mathrm{s}$

Therefore, the velocity of the particle at $\mathrm{t}=1\mathrm{s}$is $-6\mathrm{m}/\mathrm{s}$.

From the step 3, velocity is less than 0, $\mathrm{v}<0$, the particle is moving along -x direction at$\mathrm{t}=1\mathrm{s}$ .

At given time$\mathrm{t}=1\mathrm{s}$, the velocity is$-6\mathrm{m}/\mathrm{s}$ . Therefore, the magnitude of velocity, i.e. speed is$\left|\mathrm{v}\right|=6\mathrm{m}/\mathrm{s}$.

From the equation $\mathrm{v}=-12+6\mathrm{t}$, we see that at$\mathrm{t}=2\hspace{0.33em}\mathrm{s}$ velocity becomes 0 .

So, from $0\hspace{0.33em}\mathrm{s}\mathrm{to}2\hspace{0.33em}\mathrm{s}$velocity decreasing and at $\mathrm{t}=3\hspace{0.33em}\mathrm{s}$we get velocity 6 m/s that means it is increasing.

The equation for the velocity is given as,

$\mathrm{v}=-12+6\left(\mathrm{t}\right)$

Substitute the velocity equals to zero.

$$\begin{gathered} 0=-12+6\left(t\right) \\ \mathrm{t}=2\hspace{0.33em}\mathrm{s} \end{gathered}$$

Therefore, at $\mathrm{t}=2\hspace{0.33em}\mathrm{s}$ velocity becomes 0 .

Using equation (i), it can be seen that for $\mathrm{t}>2\hspace{0.33em}\mathrm{s},\mathrm{v}>0.$,

Therefore, after 3 sec, velocity will never be zero.