The acceleration of a particle moving only on a horizontalxyplane is given by $\mathbf{}\mathbf{a}\mathbf{}\mathbf{⃗}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{t}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{t}\hat{\mathbf{j}}$ , where $\mathbf{a}\mathbf{}\mathbf{⃗}\mathbf{}$ is in meters per second squared and tis in seconds. At $\mathbf{t}\mathbf{=}\mathbf{0}$ , the position vector $\mathbf{r}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\mathbf{}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\mathbf{}\hat{\mathbf{j}}$ locates the particle, which then has the velocity vector $\mathbf{}\mathbf{v}\mathbf{}\mathbf{⃗}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{}\hat{\mathbf{j}}$ . At $\mathbf{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the xaxis.

It is given that,

i Acceleration of particles in xy plane is given by $\mathrm{a}⃗=\left(3\mathrm{t}\right)\hat{\mathrm{i}}+\left(4\mathrm{t}\right)\hat{\mathrm{j}}$where $\mathrm{a}⃗$in $\mathrm{m}/{\mathrm{s}}^2$& t in s.

ii The position vector of particle is $\mathrm{r}⃗=(20.0)\hat{\mathrm{i}}+(40.0)\hat{\mathrm{j}}$at time $\mathrm{t}=0\hspace{0.33em}\mathrm{s}$

iii Velocity vector $\mathrm{v}⃗$ of particle is $\mathrm{v}⃗=(5.0)\hat{\mathrm{i}}+(2.0)\hat{\mathrm{j}}$at time$\mathrm{t}=0\hspace{0.33em}\mathrm{s}$

This problem deals with the indefinite integral which is commonly applied in problems involving distance, velocity, and acceleration, with respect to time. With this operation, the velocity vector and the position vector can be found. Applying integration over acceleration, the velocity vector can be found. Similarly, the position vector can be expressed in a unit vector notation by integrating the velocity with respect to time. Using the standard formula for the direction, the angle between the direction of travel and the positive direction of x can be found.

Formula:

The velocity is given by,

$\overrightarrow{\mathrm{v}}=\overrightarrow{{\mathrm{v}}_0}+{\int }_0^{\mathrm{t}}\stackrel{\rightharpoonup }{\mathrm{a}}\mathrm{dt}\left(\mathrm{i}\right)$

The position vector is given by,

$\overrightarrow{\mathrm{r}}=\overrightarrow{{\mathrm{r}}_0}+{\int }_0^{\mathrm{t}}\stackrel{\rightharpoonup }{\mathrm{v}}\mathrm{dt}\left(\mathrm{ii}\right)$

The angle can be written as,

$\mathrm{\theta }={\tan }^{-1}\left(\frac{{\stackrel{\rightharpoonup }{\mathrm{v}}}_{\mathrm{y}}}{\stackrel{\rightharpoonup }{{\mathrm{v}}_{\mathrm{x}}}}\right)\left(\mathrm{iii}\right)$

It is given that,

$\mathrm{a}⃗=\left(3\mathrm{t}\right)\hat{\mathrm{i}}+\left(4\mathrm{t}\right)\hat{\mathrm{j}}$

Substituting the above value in equation (i),

$$\begin{gathered} \stackrel{\rightharpoonup }{\mathrm{v}}={\stackrel{\rightharpoonup }{\mathrm{v}}}_0+{\int }_0^{\mathrm{t}}\overrightarrow{\mathrm{a}}\mathrm{dt} \\ \stackrel{\rightharpoonup }{\mathrm{v}}=\left(5.0\right)\hat{\mathrm{i}}+\left(2.0\right)\hat{\mathrm{j}}+{\int }_0^{\mathrm{t}}\left[\left(3\mathrm{t}\right)\hat{\mathrm{i}}+\left(4\mathrm{t}\right)\hat{\mathrm{j}}\right]\mathrm{dt} \\ \stackrel{\rightharpoonup }{\mathrm{v}}=(5.00+3{\mathrm{t}}^2/2)\hat{\mathrm{i}}+(2.00+2{\mathrm{t}}^2)\hat{\mathrm{j}} \end{gathered}$$

Substituting the value of $\stackrel{\rightharpoonup }{\mathrm{v}}$ in equation (ii), the position vector will be,

$$\begin{gathered} \stackrel{\rightharpoonup }{\mathrm{r}}=\left(20.0\right)\hat{\mathrm{i}}+\left(40.0\right)\hat{\mathrm{j}}+{\int }_0^{\mathrm{t}}\left[\left(5.00+\frac{3{\mathrm{t}}^2}{2}\right)\hat{\mathrm{i}}+\left(2.00+2{\mathrm{t}}^2\right)\hat{\mathrm{j}}\right]\mathrm{dt} \\ \stackrel{\rightharpoonup }{\mathrm{r}}=\left(20.0+5.00\mathrm{t}+\frac{{\mathrm{t}}^3}{2}\right)\hat{\mathrm{i}}+\left(40.0+2.00\mathrm{t}+\frac{2{\mathrm{t}}^3}{3}\right)\hat{\mathrm{j}} \end{gathered}$$

Therefore, position vector at time is,$\mathrm{t}=4.00\mathrm{s}$

$\stackrel{\rightharpoonup }{\mathrm{r}}=(72.0\mathrm{m})\hat{\mathrm{i}}+(90.7\mathrm{m})\hat{\mathrm{j}}$

It is given that,

$\stackrel{\rightharpoonup }{\mathrm{v}}\left(\mathrm{t}=4.00\mathrm{s}\right)=\left(29.0\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(34.0\right)\hat{\mathrm{j}}$

Hence,
$$\begin{gathered} \mathrm{\theta }={\tan }^{-1}\left(\frac{34.0}{29.0}\right) \\ \mathrm{\theta }=49.5^{\circ } \end{gathered}$$