Question

Question: A particle initially located at the origin has an acceleration of$\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$and an initial velocity of. ${\overrightarrow{\mathbf{v}}}_{\mathbf{i}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. Find

(a)The vector position of the particle at any time t,

(b)The velocity of the particle at any time t,

(c)The coordinates of the particle at $\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$, and

(d)The speed of the particle at $\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$.

Short answer

(a)The position vector of the particle is $5t\mathit{}\mathit{+}\mathit{}\mathit{1}\mathit{.}\mathit{5}{t}^{\mathit{2}}\mathit{}\overbrace{\mathrm{j}}$.

(b)The velocity of the particle at any time is $5\hat{\mathrm{i}}+3\mathrm{t}\hat{\mathrm{j}}$

(c)The coordinates of the particle at time t=2 is X =10 m and y = 6 m

(d)The speed of the particle is $7.81\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Definition of velocity vector and position vector.

The derivative of the displacement of particle with respect to time is called velocity. The derivative of the velocity of particle with respect to time is called acceleration. On the other hand, the integration of the acceleration with respect to time gives velocity and the integration of the velocity with respect to time gives acceleration.

Step 2:

Given,

The velocity of the particle is given as-$\stackrel{\rightharpoonup }{{\mathrm{v}}_{\mathrm{i}}}=5\hat{i}m/s$

The acceleration of the particle is given as-${\stackrel{\rightharpoonup }{\mathrm{a}}}_i=3\hat{\mathrm{j}}m/s^2$

Determine the vector position of the particle.

(a)

In order to find the position vector, we use the following equation:
$$\begin{gathered} {\overrightarrow{\mathrm{r}}}_f={\hat{r}}_i+{\overrightarrow{v}}_{i}t+\frac{1}{2}+\overrightarrow{a}{t}^2 \\ {\overrightarrow{r}}_f=5t\hat{\mathrm{i}}+1.5t^2\hat{j} \end{gathered}$$

Thus, the position vector of the particle is $5\mathrm{t}\hat{\mathrm{i}}+1.5{\mathrm{t}}^2\hat{\mathrm{j}}$.

Determine the velocity of the particle.

(b)

In order to find the velocity vector, we use the following equation:
$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_f={\overrightarrow{v}}_i+\overrightarrow{a}t \\ {\overrightarrow{v}}_f=5\hat{i}+3t\hat{\mathrm{j}} \end{gathered}$$
Thus, the velocity of the particle at any time is$5\hat{i}+3t\hat{j}$

To find the coordinates of the particle.

(c)

We can also find the velocity by differentiating with respect tot.
In order to find the particle position at, we plug into

We found in part: (a) :

$$\begin{gathered} {\overrightarrow{\mathrm{r}}}_f=5t\hat{\mathrm{i}}+1.5t^2\hat{\mathrm{j}} \\ =(5\times 2)\hat{\mathrm{i}}+(1.5\times 2)\hat{\mathrm{j}} \\ =10\hat{\mathrm{i}}+6\hat{\mathrm{j}} \end{gathered}$$

x = 10 m and y = 6 m

Thus, the coordinates of the particle at time t = 2s is x = 10 m and y = 6 m.

Determine the speed of the particle.

(d)

In order to find the particle speed at , we plug into we found in part (b):
$$\begin{gathered} {\stackrel{\rightharpoonup }{\mathrm{v}}}_f=5\hat{\mathrm{i}}+3\mathrm{t}\hat{\mathrm{j}} \\ =\left(5\right)\hat{\mathrm{i}}+(3\mathrm{x}2)\hat{\mathrm{j}} \\ =5\hat{\mathrm{i}}+6\hat{\mathrm{j}} \end{gathered}$$
The magnitude is found by:
$$\begin{gathered} \left|{\overrightarrow{v}}_f\right|=\sqrt{v_x^2+v_y^2} \\ =\sqrt{{\left(5\text{m/s}\right)}^2+{\left(6\text{m/s}\right)}^2} \\ =7.81\text{m/s} \end{gathered}$$
Therefore, the speed of the particle is 7.81 m/s .