91Ó°ÊÓ

$$\begin{gathered} x=4\mathrm{t}-3 \\ x=-4{\mathrm{t}}^2+6\mathrm{t}-3 \\ x=4{\mathrm{t}}^2+6\mathrm{t}-3 \end{gathered}$$

The problem is based on Newton’s second law of motion. The law states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object. The external force acting on the rock is directly proportional to its acceleration. Differentiate the given function with respect to time which gives the velocity of the rock. To find the acceleration of rock, differentiate the velocity function with respect to the time.

Formulae:

The net force is given by, Newton’s law,

${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{m}\stackrel{\mathbf{→}}{\mathbf{a}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

$\stackrel{\mathbf{→}}{\mathbf{a}}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}\stackrel{\mathbf{→}}{\mathbf{x}}}{{\mathbf{dt}}^{\mathbf{2}}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(2)

According to Newton’s law, given by equation (1), the force acting on the rock is directly proportional to its acceleration.

We can find the acceleration of the rock by taking the double derivative of the given function x (t) at t > 0

role="math" localid="1657002514205" $$\begin{gathered} \mathrm{For}\mathrm{x}=-4{\mathrm{t}}^2+6\mathrm{t}-3 \\ \stackrel{→}{\mathrm{a}}=\frac{{\mathrm{d}}^2}{{\mathrm{dt}}^2}\left(-4{\mathrm{t}}^2+6\mathrm{t}-3\right) \\ \stackrel{→}{\mathrm{a}}=-8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{For}\mathrm{x}=4{\mathrm{t}}^2+6\mathrm{t}-3 \\ \stackrel{→}{\mathrm{a}}=\frac{{\mathrm{d}}^2}{{\mathrm{dt}}^2}\left(4{\mathrm{t}}^2+6\mathrm{t}-3\right) \\ \stackrel{→}{\mathrm{a}}=8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

For $\mathrm{x}=-4{\mathrm{t}}^2+6\mathrm{t}-3$The acceleration is negative. Hence the force $\stackrel{→}{F}$is negative.

The function of $\stackrel{→}{F}$which is directed opposite the rock’s initial direction of motion is, $\mathrm{x}=-4{\mathrm{t}}^2+6\mathrm{t}-3$