Chapter 11.5, Problem 11.141P

Race car A is traveling on a straight portion of the track while race car B is traveling on a circular portion of the track. At the instant shown, the speed of A is increasing at the rate of 10 m/s², and the speed of B is decreasing at the rate of 6 m/s². For the position shown, determine (a) the velocity of B relative to A, (b) the acceleration of B relative to A.

Fig. P11.141

To determine

The relative velocity $\left(v_{B/A}\right)$ of point B with respective to point A.

Answer to Problem 11.141P

The relative velocity $\left(v_{B/A}\right)$ of point B with respective to point A is $\underset{\_}{189.5\text{km/h}}$ at an angle $\left(\beta \right)$ of $\underset{\_}{54°}$.

Explanation of Solution

Given Information:

The speed $\left(v_A\right)$ of racer A is $240\text{km/h}$.

The speed $\left(v_B\right)$ of racer B is $200\text{km/h}$ at an angle $50°$.

The speed ${\left(a_A\right)}_t$ of car A increasing in the rate of $10{\text{m/s}}^{\text{2}}$.

The speed ${\left(a_A\right)}_t$ of car B decreasing in the rate of $6{\text{m/s}}^{\text{2}}$.

The radius of the circular path $\left(\rho \right)$ is $300\text{m}$

Calculation:

Convert unit kilometer per hour to meter per second.

Consider the velocity $\left(v_A\right)$.

$\begin{array}{c}{v}_A=240\text{km/h}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{hr}}{3600\sec }\\ =66.67\text{m/s}\end{array}$

Consider the velocity $\left(v_B\right)$.

$\begin{array}{c}{v}_B=200\text{km/h}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{hr}}{3600\sec }\\ =55.56\text{m/s}\end{array}$

Write the velocity of B$\left(v_B\right)$ in term of vector notation:

$\begin{array}{c}{v}_B=55.56\cos 50°i+55.56\sin 50°j\\ =35.71i+42.56j\end{array}$

Write the velocity of A$\left(v_A\right)$ in term of vector notation:

$v_A=66.67i$

Calculate the relative velocity vector $\left(v_{B/A}\right)$ for point B to A:

$v_{B/A}=v_B-v_A$

Substitute $66.67i$ for $v_A$ and $35.71i+42.56j$ for $v_B$.

$\begin{array}{c}{v}_{B/A}=35.71i+42.56j-66.67i\\ =-30.96i+42.56j\end{array}$

Here, ${\left(v_{B/A}\right)}_x$ is $-30.96$ and ${\left(v_{B/A}\right)}_y$ is $42.56$

Calculate the relative velocity $\left(v_{B/A}\right)$ at point B to A using the relation:

$v_{B/A}=\sqrt{{\left(v_{B/A}\right)}_x^2+{\left(v_{B/A}\right)}_y^2}$

Substitute $-30.96$ for ${\left(v_{B/A}\right)}_x$ and $42.56$ for ${\left(v_{B/A}\right)}_y$.

$\begin{array}{c}{v}_{B/A}=\sqrt{{\left(-30.96\right)}^2+{\left(42.56\right)}^2}\\ =\sqrt{2769.8752}\\ =52.63\text{m/s}\times \frac{1\text{km}}{1000\text{m}}\times \frac{3600\sec }{1\text{hr}}\\ =189.5\text{km/h}\end{array}$

Calculate the angle $\left(\beta \right)$ using the relation:

$\beta ={\tan }^{-1}\left(\frac{{\left(v_{B/A}\right)}_y}{{\left(v_{B/A}\right)}_x}\right)$

Substitute $-30.96$ for ${\left(v_{B/A}\right)}_x$ and $42.56$ for ${\left(v_{B/A}\right)}_y$.

$\begin{array}{c}\beta ={\tan }^{-1}\left(\frac{42.56}{-30.96}\right)\\ =53.966°\\ \approx 54°\end{array}$

Therefore, the relative velocity $\left(v_{B/A}\right)$ of point B with respective to point A is $\underset{\_}{189.5\text{km/h}}$ at an angle $\left(\beta \right)$ of $\underset{\_}{54°}$.

To determine

The relative acceleration $\left(a_{B/A}\right)$ of point B with respective to point A.

Answer to Problem 11.141P

The relative acceleration $\left(a_{B/A}\right)$ of point B with respective to point A is $\underset{\_}{21.8{\text{m/s}}^{\text{2}}}$ at an angle $\left(\alpha \right)$ of $\underset{\_}{5.3°}$.

Explanation of Solution

Given Information:

The speed $\left(v_A\right)$ of racer A is $240\text{km/h}$.

The speed $\left(v_B\right)$ of racer B is $200\text{km/h}$ at an angle $50°$.

The speed ${\left(a_A\right)}_t$ of car A increasing in the rate of $10{\text{m/s}}^{\text{2}}$.

The speed ${\left(a_A\right)}_t$ of car B decreasing in the rate of $6{\text{m/s}}^{\text{2}}$.

The radius of the circular path $\left(\rho \right)$ is $300\text{m}$

Calculation:

Calculate the normal acceleration ${\left(a_B\right)}_n$ at point B using the relation:

${\left(a_B\right)}_n=\frac{v^2}{\rho }$

Substitute $55.67\text{m/s}$ for v and 300 m for $\rho$.

$\begin{array}{c}{\left(a_B\right)}_n=\frac{{\left(55.56\right)}^2}{300}\\ =10.28{\text{m/s}}^{\text{2}}\end{array}$

The normal acceleration at an angle:

$\begin{array}{c}\varphi =90°-50°\\ =40°\end{array}$

Write the normal acceleration of B ${\left(a_B\right)}_n$ in term of vector notation:

${\left(a_B\right)}_n=\left({\left(a_B\right)}_n\cos \varphi \right)i-\left({\left(a_B\right)}_n\sin \varphi \right)j$

Substitute $10.28{\text{m/s}}^{\text{2}}$ for ${\left(a_B\right)}_n$ and $40°$ for $\varphi$.

$\begin{array}{c}{\left(a_B\right)}_n=10.28\cos 40°i-10.28\sin 40°j\\ =7.875i-6.608j\end{array}$

Write the tangential acceleration ${\left(a_B\right)}_t$ in term of vector notation:

${\left(a_B\right)}_t=\left({\left(a_B\right)}_t\cos 50°\right)i+\left({\left(a_B\right)}_t\sin 50°\right)j$

Substitute $6{\text{m/s}}^{\text{2}}$ for ${\left(a_B\right)}_t$ and $40°$ for $\varphi$.

$\begin{array}{c}{\left(a_B\right)}_t=\left(6\cos 50°\right)i+\left(6\sin 50°\right)j\\ =3.857i+4.596j\end{array}$

Write the tangential acceleration ${\left(a_A\right)}_t$ in term of vector notation:

${\left(a_A\right)}_t=-10i$

The normal acceleration ${\left(a_A\right)}_n$ at point A is 0 because the $\left(\rho \right)$ is infinite.

Calculate the relative acceleration vector $\left(a_{B/A}\right)$ of point B with respective to point A using the relation:

$\left(a_{B/A}\right)=a_B-a_A$

Rewrite the above equation.

$\left(a_{B/A}\right)=\left({\left(a_B\right)}_t+{\left(a_B\right)}_n\right)-\left({\left(a_A\right)}_t+{\left(a_A\right)}_n\right)$

Substitute $3.857i+4.596j$ for ${\left(a_B\right)}_t$, $7.875i-6.608j$ for ${\left(a_B\right)}_n$, $-10i$ for ${\left(a_A\right)}_t$ and 0 for ${\left(a_A\right)}_n$.

$\begin{array}{c}\left(a_{B/A}\right)=\left(3.857i+4.596j\right)+\left(7.875i-6.608j\right)-\left(-10i+0\right)\\ =21.732i-2.012j\end{array}$

Here, ${\left(a_{B/A}\right)}_x$ is $21.732{\text{m/s}}^{\text{2}}$ and ${\left(a_{B/A}\right)}_y$ is $-2.012{\text{m/s}}^{\text{2}}$.

Calculate the relative acceleration $\left(v_{B/A}\right)$ of point B with respective to point A using the relation:

$a_{B/A}=\sqrt{{\left(a_{B/A}\right)}_x^2+{\left(a_{B/A}\right)}_y^2}$

Substitute, $21.732$ for ${\left(a_{B/A}\right)}_x$ and $-2.012$ for ${\left(a_{B/A}\right)}_y$.

$\begin{array}{c}{a}_{B/A}=\sqrt{{21.732}^2+{\left(-2.012\right)}^2}\\ =\sqrt{476.327}\\ =21.8{\text{m/s}}^{\text{2}}\end{array}$

Calculate the angle $\left(\alpha \right)$ using the relation:

$\alpha ={\tan }^{-1}\left(\frac{{\left(a_{B/A}\right)}_y}{{\left(a_{B/A}\right)}_x}\right)$

Substitute, $21.732$ for ${\left(a_{B/A}\right)}_x$ and $-2.012$ for ${\left(a_{B/A}\right)}_y$.

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{-2.012}{21.732}\right)\\ =5.3°\end{array}$

Therefore, the relative acceleration $\left(a_{B/A}\right)$ of point B with respective to point A is $\underset{\_}{21.8{\text{m/s}}^{\text{2}}}$ at an angle $\left(\alpha \right)$ of $\underset{\_}{5.3°}$.