Answer
$|A_{B/A}| = 9.167\frac{m}{s^{s}}$
$\theta_{A} =74.84$ degrees (third quadrant)
$|V_{B/A}| = 20.53\frac{m}{s}$
$\theta_{V} = 43.06$ degrees (third quadrant)
Work Step by Step
First, define the velocities of cars B and A in terms of $i$ and $j$ components:
$V_{B}=-30sin(30)i+30cos(30)j$
$V_{A} = 40j$
The question wants us to find the velocity of car B with respect to car A:
$V_{B/A}=V_{B}-V_{A}$
$V_{B/A}=-30sin(30)i+30cos(30)j-40j$
$V_{B/A}=-15i-14.0192j$
$|V_{B/A}| = \sqrt {(-15)^2+(-14.0192)^2} = 20.53\frac{m}{s}$
Calculate $a_{n}$ and $a_{t}$ for car B:
$a_{n} = \frac{v_{b}^{2}}{\rho} =\frac{30^{2}}{200} =4.5\frac{m}{s^{2}}$
$a_{t} = -3\frac{m}{s^{2}}$
Define the acceleration of cars B and A in term of $i$ and $j$ components:
$A_{B}=-(-3)sin(30)i-4.5sin(60)i+(-3)cos(30)j-(4.5)cos(60)j$
$A_{A} = 4j$
The question wants us to find the acceleration of car B with respect to car A:
$A_{B/A}=A_{B}-A_{A}$
$A_{B/A}=-2.397i-4.848j - 4j$
$A_{B/A}=-2.397i-8.848j$
$|A_{B/A}| = \sqrt {(-2.397)^2+(-8.848)^2} = 9.167\frac{m}{s^{s}}$
Lastly, calculate the angle of inclination of $V_{B/A}$ and $A_{B/A}$:
$\theta_{V} = arctan(\frac{-14.0192}{-15})=43.06$ degrees
Since the x and y components are both negative, this angle is located in the third quadrant.
$\theta_{A} = arctan(\frac{-8.848}{-2.397})=74.84$ degrees
Since the x and y components are both negative, this angle is located in the third quadrant.
