Question

The position vector $\overrightarrow{\mathbf{r}}$of a particle moving in the xy plane is $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{2}\mathbf{t}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{sin}\left[\left(\mathrm{\tau \tau }/4\mathrm{rad}/s\right)t\right]\hat{\mathbf{j}}$ , with in meters and tin seconds. (a) Calculate the x and y components of the particle’s position at $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{,}$and 4.0 sand sketch the particle’s path in the xy plane for the interval $\mathbf{0}\mathbf{\le }\mathit{t}\mathbf{\le }\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$. (b) Calculate the components of the particle’s velocity at t = 1.0, 2.0, and 3.0 s . Show that the velocity is tangent to the path of the particle and in the direction the particle is moving at each time by drawing the velocity vectors on the plot of the particle’s path in part (a). (c) Calculate the components of the particle’s acceleration at t =1.0 , 2.0, and 3.0 s .

Short answer

a) x and y components of the particle’s position (m) are,

$$\begin{gathered} r_{x0}=0,r_{x1}=2,r_{x2}=4,r_{x3}=6,r_{x4}=8; \\ {r}_{y0}=0,r_{y1}=1.4,r_{y2}=2,r_{y3}=1.4,r_{y4}=0; \end{gathered}$$

b) and components of the particle’s velocity

$$\begin{gathered} v_{x1}=2,v_{x2}=2,v_{x3}=2, \\ {v}_{y1}=1.1,v_{y2}=2,v_{y3}=-1.1, \end{gathered}$$

c) and components of the particle’s acceleration

$$\begin{gathered} a_{x1}=0,a_{x2}=0,a_{x3}=0 \\ {a}_{y1}=-0.87,a_{y2}=-1.2,a_{y3}=-0.87 \end{gathered}$$.

Step-by-step solution

Given information

It is given that,

Position vector$r\left(t\right)=\left(2t\right)\hat{i}+2\sin \left(\frac{\pi t}{4}\right)\hat{j}$

Determining the concept

This problem is based on the resolution of components of vector. The resolution of a vector is the splitting of a single vector into two or more vectors in different directions. These vectors together which together exhibit similar properties as is exhibited by a single vector itself. The velocity and acceleration vector can be found by differentiating the position vector with respect to t.

Formula:

If r (t) is a position vector then, velocity vector is given by,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$ (i)

Acceleration vector is given by,

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$ (ii)

(a) determining the x and y components of the particle’s position

x And y components of the particle’s position (m) are,

$$\begin{gathered} r_x=2t \\ {r}_y=2\sin \left(\frac{\pi t}{4}\right) \end{gathered}$$

Hence,

$$\begin{gathered} r_{x0}=0,r_{x1}=2,r_{x2}=4,r_{x3}=6,r_{x4}=8; \\ {r}_{y0}=0,r_{y1}=1.4,r_{y2}=2,r_{y3}=1.4,r_{y4}=0; \end{gathered}$$

Sketch of particle’s path in plane is given below:

(b) determining the x and y components of the particle’s velocity

x And y components of the particle’s velocity (m/s) are,

Differentiating equation (i),

$$\begin{gathered} \overrightarrow{v}=2\hat{i}+\frac{\mathrm{\pi }}{2}cos\left(\frac{\pi t}{4}\right)\hat{j} \\ {v}_x=2 \\ {v}_y=\frac{\pi }{2}\cos \left(\frac{\pi t}{4}\right)\left(\mathrm{iii}\right) \end{gathered}$$

And components of the particle’s velocity (m/s) are,

$$\begin{gathered} v_{x1}=2,v_{x2}=2,v_{x3}=2, \\ {v}_{y1}=1.1,v_{y2}=0,v_{y3}=-1.1 \end{gathered}$$

(c) determining the x and y components of the particle’s acceleration

x And y components of the particle’s acceleration $\left(\mathrm{m}/{\mathrm{s}}^2\right)$are,

Differentiating equation (iii),

$\overrightarrow{a}=\frac{-{\pi }^2}{8}\sin \left(\frac{\pi t}{4}\right)$

Hence, x and y components of the particle’s acceleration $\left(\mathrm{m}/{\mathrm{s}}^2\right)$are,

$$\begin{gathered} a_{x1}=0,a_{x2}=0,a_{x3}=0 \\ {a}_{y1}=-0.87,a_{y2}=-1.2,a_{y3}=-0.87 \end{gathered}$$