Chapter 3, Problem 51P

To determine

The velocity of the particle

Answer to Problem 51P

The velocity of the particle is $44.72\text{m/s}$ $26.56°$ south-east direction.

Explanation of Solution

Write the expression for the acceleration component due south.

    $a_{\text{south}}=a\cos \theta$        (I)

Here, $a_{\text{south}}$ is the acceleration in the south direction, $a$ is the magnitude of acceleration and $\theta$ is the angle that velocity makes with the horizontal component.

Write the expression for the acceleration component due to east.

    $a_{\text{east}}=a\sin \theta$        (II)

Here, $a_{\text{east}}$ is the acceleration in the east direction, $a$ is the magnitude of acceleration and $\theta$ is the angle that velocity makes with the vertical component.

Write the expression for the velocity at east direction after $8$ seconds.

    $v_{\text{east}}=v_i+a_{\text{east}}t$        (III)

Here, $v_{\text{east}}$ is the velocity in the east direction, $v_i$ is the initial velocity, $a_{east}$ is the acceleration in the east direction and $t$ is the time.

Write the expression for the velocity at south direction after $8$ seconds.

    $v_{\text{south}}=v_i+a_{\text{south}}t$        (IV)

Here, $v_{\text{south}}$ is the velocity in the south direction, $v_i$ is the initial velocity, $a_{\text{south}}$ is the acceleration in the south direction and $t$ is the time.

Write the equation for the speed after $8$ seconds.

    $v=\sqrt{{\left(v_{\text{east}}\right)}^2+{\left(v_{\text{south}}\right)}^2}$        (V)

Here, $v$ is the speed after $8$ seconds, $v_{\text{east}}$ is the velocity in the east direction and $v_{\text{south}}$ is the velocity in the south direction.

Write the expression to find the direction of the velocity,

  $\theta ={\tan }^{-1}\left(\frac{v_{\text{south}}}{v_{\text{north}}}\right)$        (VI)

Here, $\theta$ is the direction, $v_{\text{south}}$ is the velocity in south, and $v_{\text{north}}$ is velocity in north

Conclusion;

Substitute $2.50{\text{m/s}}^{\text{2}}$ for $a$ and $0°$ for $\theta$ in equation (I).

    $\begin{array}{c}{a}_{\text{south}}=\left(2.50{\text{m/s}}^{\text{2}}\right)\cos 0°\\ =2.50{\text{m/s}}^{\text{2}}\end{array}$

Substitute $2.50{\text{m/s}}^{\text{2}}$ for $a$ and $0°$ for $\theta$ in equation (II).

    $\begin{array}{c}{a}_{\text{east}}=\left(2.50{\text{m/s}}^{\text{2}}\right)\sin 0°\\ =0\end{array}$

Substitute $40\text{m/s}$ for $v_i$, $0$ for $a_{\text{east}}$ in equation (III).

    $\begin{array}{c}{v}_{\text{east}}=40\text{m/s}+\left(0\right)t\\ =40\text{m/s}\end{array}$

Substitute $0\text{m/s}$ for $v_i$, $2.50{\text{m/s}}^{\text{2}}$ for $a_{\text{south}}$ and $8\text{s}$ for $t$ in equation (IV).

    $\begin{array}{c}{v}_{\text{south}}=0\text{m/s}+\left(2.50{\text{m/s}}^{\text{2}}\right)\left(8\text{s}\right)\\ =20\text{m/s}\end{array}$

Substitute $40\text{m/s}$ for $v_{\text{east}}$ and $20\text{m/s}$ for $v_{\text{south}}$ in equation (V).

    $\begin{array}{c}v=\sqrt{{\left(40\text{m/s}\right)}^2+{\left(20\text{m/s}\right)}^2}\\ =44.72\text{m/s}\end{array}$

Substitute $40\text{m/s}$ for $v_{\text{east}}$ and $20\text{m/s}$ for $v_{\text{south}}$ in equation (VI).

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{20\text{m/s}}{4\text{0m/s}}\right)\\ =26.56°\end{array}$

Therefore, the velocity of the particle is $44.72\text{m/s}$  $26.56°$ south-east direction.

$44.72\text{m/s}$.