Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at $t=2\text{}\text{ s}$ .

Answer to Problem 50P

  $\overrightarrow{v}\left(2\right)=10\widehat{i}-3\widehat{j}$

Explanation of Solution

Given data

  $\begin{array}{l}x\left(0\right)=4.0\text{m}\\ y\left(0\right)=3.0\text{m}\\ \overrightarrow{v}\left(0\right)=2\widehat{i}-9\widehat{j}\\ \overrightarrow{a}\left(t\right)=4\widehat{i}+3\widehat{j}\end{array}$

Formula used

  $\overrightarrow{v}\left(t\right)={\displaystyle \int \overrightarrow{a}\left(t\right)}dt$

Calculation

  $\begin{array}{l}\overrightarrow{v}\left(t\right)={\displaystyle \int \overrightarrow{a}\left(t\right)}dt\\ \overrightarrow{v}\left(t\right)={\displaystyle \int \left(4\widehat{i}+3\widehat{j}\right)}dt\end{array}$

  $\overrightarrow{v}\left(t\right)=4\widehat{i}\left(t+a\right)+3\widehat{j}\left(t+b\right)$ where a and b are constants.

Substitute the values for $\overrightarrow{v}\left(t\right)$ when $t=0\text{}\text{ s}$

  $\overrightarrow{v}\left(0\right)=4\widehat{i}\left(a\right)+3\widehat{j}\left(b\right)=2\widehat{i}-9\widehat{j}$

Comparing values of $\widehat{i}$ ,

  $\begin{array}{l}4a=2\\ a=0.5\end{array}$

Comparing values of $\widehat{j}$ ,

  $\begin{array}{l}3b=-9\\ b=-3\end{array}$

Substituting values of a and b,

  $\overrightarrow{v}\left(t\right)=4\left(t+0.5\right)\widehat{i}+3\left(t-3\right)\widehat{j}$

when $t=2\text{}\text{ s}$

  $\begin{array}{l}\overrightarrow{v}\left(2\right)=4\left(2+0.5\right)\widehat{i}+3\left(2-3\right)\widehat{j}\\ \overrightarrow{v}\left(2\right)=4\left(2.5\right)\widehat{i}+3\left(-1\right)\widehat{j}\\ \overrightarrow{v}\left(2\right)=10\widehat{i}-3\widehat{j}\end{array}$

Conclusion

  $\overrightarrow{v}\left(2\right)=10\widehat{i}-3\widehat{j}$

(b)

To determine

To Calculate: The position vector at $t=4\text{}\text{ s}$.

Answer to Problem 50P

  $\overrightarrow{s}\left(4\right)=41\widehat{i}-3\widehat{j}$

Magnitude of the displacement at $t=4\text{}\text{ s}$= $\left|\overrightarrow{s}\left(4\right)\right|$

  $\left|\overrightarrow{s}\left(4\right)\right|=\sqrt{{41}^2+{\left(-3\right)}^2}=\sqrt{1681+9}=\sqrt{1690}=6.411\text{ m}$

Direction of the displacement at $t=4\text{}\text{ s}$= $\theta$

  $\theta ={\tan }^{-1}\left(\frac{3}{41}\right)={\tan }^{-1}\left(0.0732\right)={4.19}^{\circ }$ from positive x direction to negative y direction.

Explanation of Solution

Given data

  $\overrightarrow{v}\left(t\right)=4\widehat{i}\left(t+0.5\right)+3\widehat{j}\left(t-3\right)$

Formula used

  $\overrightarrow{s}\left(t\right)={\displaystyle \int \overrightarrow{v}\left(t\right)}dt$

Calculation

  $\begin{array}{l}\overrightarrow{s}\left(t\right)={\displaystyle \int \overrightarrow{v}\left(t\right)}dt\\ \overrightarrow{s}\left(t\right)={\displaystyle \int \left[4\widehat{i}\left(t+0.5\right)+3\widehat{j}\left(t-3\right)\right]}dt\end{array}$

  $\overrightarrow{s}\left(t\right)=4\widehat{i}\left(\frac{t^2}{2}+0.5t+c\right)+3\widehat{j}\left(\frac{t^2}{2}-3t+d\right)$ where c and d are constants.

Substituting values for $\overrightarrow{s}\left(t\right)$ when $t=0\text{}\text{ s}$

  $\begin{array}{l}\overrightarrow{s}\left(0\right)=4\left(\frac{0^2}{2}+0.5\times 0+c\right)\widehat{i}+3\left(\frac{0^2}{2}-3\times 0+d\right)\widehat{j}\\ \overrightarrow{s}\left(0\right)=4c\widehat{i}+3d\widehat{j}\end{array}$

Comparing values of $\widehat{i}$

  $t=0\text{}\text{ s}$ ,

  $\begin{array}{l}4c=1\\ c=0.25\end{array}$

Comparing values of $\widehat{j}$ when $t=0\text{}\text{ s}$ ,

  $\begin{array}{l}3d=3\\ d=1\end{array}$

Substituting values of c and d,

  $\overrightarrow{s}\left(t\right)=4\left(\frac{t^2}{2}+0.5t+0.25\right)\widehat{i}+3\left(\frac{t^2}{2}-3t+3\right)\widehat{j}$

When $t=4\text{}\text{ s}$

  $\begin{array}{l}\overrightarrow{s}\left(4\right)=4\left(\frac{4^2}{2}+0.5\times 4+0.25\right)\widehat{i}+3\left(\frac{4^2}{2}-3\times 4+3\right)\widehat{j}\\ \overrightarrow{s}\left(4\right)=4\left(\frac{16}{2}+2+0.25\right)\widehat{i}+3\left(\frac{16}{2}-12+3\right)\widehat{j}\\ \overrightarrow{s}\left(4\right)=4\left(10.25\right)\widehat{i}+3\left(-1\right)\widehat{j}\\ \overrightarrow{s}\left(4\right)=41\widehat{i}-3\widehat{j}\end{array}$

  $\left|\overrightarrow{s}\left(4\right)\right|=\sqrt{{41}^2+{\left(-3\right)}^2}=\sqrt{1681+9}=\sqrt{1690}=6.411\text{ m}$

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{3}{41}\right)={\tan }^{-1}\left(0.0732\right)={4.19}^{\circ }\\ \end{array}$

Conclusion

  $\overrightarrow{s}\left(4\right)=41\widehat{i}-3\widehat{j}$

The magnitude of the displacement at $t=4\text{}\text{ s}$= $\left|\overrightarrow{s}\left(4\right)\right|$

  $\left|\overrightarrow{s}\left(4\right)\right|=\sqrt{{41}^2+{\left(-3\right)}^2}=\sqrt{1681+9}=\sqrt{1690}=6.411\text{ m}$

Direction of the displacement at $t=4\text{}\text{ s}$ = $\theta$

  $\theta ={\tan }^{-1}\left(\frac{3}{41}\right)={\tan }^{-1}\left(0.0732\right)={4.19}^{\circ }$ from positive x-direction to negative y-direction.