91Ó°ÊÓ

It is given that,

$\stackrel{→}{\mathrm{r}}=(2{\mathrm{t}}^3-5\mathrm{t})\hat{\mathrm{i}}+(6-7{\mathrm{t}}^4)\hat{\mathrm{j}}$

This problem deals with simple algebraic operation that involves calculation of position vector, velocity vector and the acceleration at given time. Velocity is the rate of change of its position and the acceleration is the rate of change of velocity with respect to time. Thus the velocity vector and acceleration can be found using the standard equation for the velocity and acceleration.

Formulae

$\stackrel{→}{v}=\frac{d\stackrel{→}{r}}{dt}$ (1)

Where, $\stackrel{→}{r}$is the position vector

$\stackrel{→}{a}=\frac{d\stackrel{→}{v}}{dt}$ (2)

The direction of the velocity vector is

$\tan \mathrm{θ}=\frac{y}{x}$ (3)

$\stackrel{→}{r}=(2t^2-5t)\hat{i}+(6-7t^4)\hat{j}$

Substituting t = 2 s in the above equation, the displacement vector will be,

$$\begin{gathered} \hat{\mathrm{r}}=(2×2^3-5×2)\hat{\mathrm{i}}+(6-7×2^4)\hat{\mathrm{j}} \\ \mathrm{Thus}, \\ \stackrel{→}{\mathrm{r}}=(6\hat{\mathrm{i}}-106\hat{\mathrm{j}})\mathrm{m} \end{gathered}$$

Thus,

Substituting $\stackrel{→}{r}$derived from part (a) in equation (i), the velocity can be written as,

$$\begin{gathered} \stackrel{→}{\mathrm{v}}=\frac{\mathrm{d}\left(\right(2{\mathrm{t}}^3-5\mathrm{t})\hat{\mathrm{i}}+(6-7{\mathrm{t}}^4\left)\hat{\mathrm{j}}\right)}{\mathrm{dt}} \\ \stackrel{→}{\mathrm{v}}=(6{\mathrm{t}}^2-5)\hat{\mathrm{i}}-25{\mathrm{t}}^3\hat{\mathrm{j}} \\ \stackrel{→}{\mathrm{v}}=(6×2^2-5)\hat{\mathrm{i}}-28×2^3\hat{\mathrm{j}} \\ \mathrm{Thus}, \\ \stackrel{→}{\mathrm{v}}=(19\hat{\mathrm{i}}-224\hat{\mathrm{j}})\mathrm{m}/\mathrm{s} \end{gathered}$$

Substituting $\stackrel{→}{v}$derived from part (b) in equation (ii), the acceleration can be written as,

$$\begin{gathered} \stackrel{→}{\mathrm{a}}=\left(12\mathrm{t}\right)\hat{\mathrm{i}}-\left(84{\mathrm{t}}^2\right)\hat{\mathrm{j}} \\ \mathrm{For}\mathrm{t}=2.00\mathrm{s} \\ \stackrel{→}{\mathrm{a}}=(12×2)\hat{\mathrm{i}}-(84×2^2)\hat{\mathrm{j}} \\ \mathrm{Thus}, \\ \stackrel{→}{\mathrm{a}}=(24\hat{\mathrm{i}}-336\hat{\mathrm{j}})\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Substituting y components from velocity vector in equation (iii) the direction will be,

$$\begin{gathered} \mathrm{³Ù²¹²Ôθ}=\frac{-224}{19} \\ \mathrm{θ}=-85.2^{°} \end{gathered}$$

An angle is negative, that means that the angle is measured in the clockwise direction below the positive x axis.