Chapter 3, Problem 3P

A particle initially located at the origin has an acceleration of $\stackrel{\to }{a}=3.00\hat{j}{\text{m/s}}^2$ and an initial velocity of ${\stackrel{\to }{v}}_i=5.00\hat{i}\text{m/s}$. 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 t = 2.00 s, and (d) the speed of the particle at t = 2.00 s.

To determine

The vector position of the particle at time $t$ .

Answer to Problem 3P

The vector position of the particle at time $t$ is $\underset{\_}{5.00t\widehat{\text{i}}+1.50t^2\widehat{\text{j}}}$.

Explanation of Solution

Write the expression for the vector position of the particle.

  ${\overrightarrow{\text{r}}}_f={\overrightarrow{\text{r}}}_i+{\overrightarrow{\text{v}}}_{i}t+\frac{1}{2}\overrightarrow{\text{a}}{t}^2$        (I)

Here, ${\overrightarrow{\text{r}}}_f$ is the final vector position of the particle, ${\overrightarrow{\text{r}}}_i$ is the initial vector position of the particle, ${\overrightarrow{\text{v}}}_i$ is the initial velocity of the particle, $\overrightarrow{\text{a}}$ is the acceleration of the particle and $t$ is the time taken.

Conclusion:

Substitute $3.00\widehat{j}{\text{m/s}}^2$ for $\overrightarrow{\text{a}}$, $5.00\widehat{\text{i}}\text{ m/s}$ for ${\overrightarrow{\text{v}}}_i$ and $0\widehat{\text{i}}+0\widehat{\text{j}}$ for ${\overrightarrow{\text{r}}}_i$ in above equation to find ${\overrightarrow{\text{r}}}_f$,

  $\begin{array}{c}{\overrightarrow{\text{r}}}_f=\left(0\widehat{\text{i}}+0\widehat{\text{j}}\right)+\left(5.00\widehat{\text{i}}\text{ m/s}\right)t+\frac{1}{2}\left(3.00\widehat{j}{\text{m/s}}^2\right)t^2\\ =5.00t\widehat{\text{i}}\text{+}\text{1}\text{.50}{t}^2\widehat{\text{j}}\end{array}$

Therefore, the vector position of the particle at time $t$ is $\underset{\_}{5.00t\widehat{\text{i}}+1.50t^2\widehat{\text{j}}}$.

To determine

The velocity of the particle at any time $t$.

Answer to Problem 3P

The velocity of the particle at any time $t$ is $5.00t\widehat{\text{i}}+3.00t\widehat{\text{j}}$.

Explanation of Solution

Write the expression for the velocity of the particle at any time $t$.

  ${\overrightarrow{\text{v}}}_f={\overrightarrow{\text{v}}}_i+\overrightarrow{\text{a}}t$        (II)

Here, ${\overrightarrow{\text{v}}}_f$ is the velocity of the particle at any time $t$, ${\overrightarrow{\text{v}}}_i$ is the initial velocity of the particle.

Conclusion:

Substitute $3.00\widehat{j}{\text{m/s}}^2$ for $\overrightarrow{\text{a}}$ and $5.00\widehat{\text{i}}\text{ m/s}$ for ${\overrightarrow{\text{v}}}_i$ in above equation to find ${\overrightarrow{\text{v}}}_f$,

  ${\overrightarrow{\text{v}}}_f=5.00\widehat{\text{i}}+3.00t\widehat{\text{j}}$

Therefore, the velocity of the particle at any time $t$ is $5.00t\widehat{\text{i}}+3.00t\widehat{\text{j}}$.

To determine

The coordinates of the particle at $t=2.00\text{s}$.

Answer to Problem 3P

The coordinates of the particle at $t=2.00\text{s}$ is $x_f=10.0\text{m, }{y}_f=6.00\text{m}$.

Explanation of Solution

Write the expression for the vector position of the particle at time $t$.

  ${\overrightarrow{\text{r}}}_f=x_f\widehat{\text{i}}+y_f\widehat{\text{j}}$        (III)

Here, $x_f$ is the $x$ component and $y_f$ is the $y$ component.

Conclusion:

Substitute $2.0\text{s}$ for $t$ a in $r_f=5.00t\widehat{\text{i}}+1.50t^2\widehat{\text{j}}$ equation to find $x_f$ and $y_f$.

  $\begin{array}{c}{\overrightarrow{\text{r}}}_f=\left(5.00\right)\left(2.00\text{s}\right)\widehat{\text{i}}+\left(1.50\right){\left(2.00\text{s}\right)}^2\widehat{\text{j}}\\ \text{=10}\text{.00}\widehat{\text{i}}+6.00\widehat{\text{j}}\end{array}$

So compare the above value with (III) to get $x_f$ and $y_f$,

  $x_f=10.0\text{m}$

  $y_f=6.0\text{m}$

Therefore, the coordinates of the particle at $t=2.00\text{s}$ is $x_f=10.0\text{m, }{y}_f=6.00\text{m}$.

To determine

The speed of the particle at $t=2.00\text{s}$.

Answer to Problem 3P

The speed of the particle at $t=2.00\text{s}$ is $7.81\text{m/s}$.

Explanation of Solution

Write the expression for the speed of the particle at time $t$.

  ${\overrightarrow{\text{v}}}_f={\overrightarrow{\text{v}}}_i+\overrightarrow{\text{a}}t$        (IV)

Write the final speed of the particle.

    $v_f=\sqrt{v_{xf}^2+v_{yf}^2}$        (V)

Here, $v_{xf}$ is the speed at $x$ component and $v_{yf}$ is the speed at $y$ component.

Conclusion:

Substitute $3.00\widehat{j}{\text{m/s}}^2$ for $\overrightarrow{\text{a}}$ and $5.00\widehat{\text{i}}\text{ m/s}$ for ${\overrightarrow{\text{v}}}_i$ in (IV) to find ${\overrightarrow{\text{v}}}_f$,

  $\begin{array}{c}{\overrightarrow{\text{v}}}_f=\left(5.00\text{m/s}\right)\widehat{\text{i}}+\left(3.00\text{m/s}\right)\left(2.00\text{s}\right)\widehat{\text{j}}\\ \text{=}\left(\text{5}\text{.00}\widehat{\text{i}}+6.00\widehat{\text{j}}\right)\text{m/s}\end{array}$

Substitute $5.00\widehat{\text{i}}\text{m/s}$ for $v_{xf}$ and $6.00\widehat{\text{j}}\text{ m/s}$ for $v_{yf}$ in (V) to find ${\overrightarrow{\text{v}}}_f$,

    $\begin{array}{c}{v}_f=\sqrt{{\left(5.00\widehat{\text{i}}\text{m/s}\right)}^2+{\left(6.00\widehat{\text{j}}\text{ m/s}\right)}^2}\\ =7.81\text{m/s}\end{array}$

Therefore, the speed of the particle at $t=2.00\text{s}$ is $7.81\text{m/s}$.