Question 11.SP.14: Automobile A is traveling east at the constant speed of 36 k...
Automobile A is traveling east at the constant speed of 36 km/h. As automobile A crosses the intersection shown, automobile B starts from rest 35 m north of the intersection and moves south with a constant acceleration of 1.2 m/s². Determine the position, velocity, and acceleration of B relative to A 5 s after A crosses the intersection.
STRATEGY: This is a relative motion problem. Determine the motion of each vehicle independently, and then use the definition of relative motion to determine the desired quantities.
MODELING and ANALYSIS:
Motion of Automobile A. Choose x and y axes with the origin at
the intersection of the two streets and with positive senses directed east
and north, respectively. First express the speed in m/s, as
The motion of A is uniform, so for any time t
a_{A}=0\\v_A=+10 \mathrm{~m} / \mathrm{s}\\x_{A}={(x_A)}_0+v_At=0+10 tFor t = 5 s, you have (Fig. 1)
Motion of Automobile B. The motion of B is uniformly acceler-ated, so
\begin{aligned}&a_{B}=-1.2 \mathrm{~m} / \mathrm{s}^{2} \\&v_{B}=\left(v_{B}\right)_{0}+a t=0-1.2 t \\&y_{B}=\left(y_{B}\right)_{0}+\left(v_{B}\right)_{0} t+\frac{1}{2} a_{B} t^{2}=35+0-\frac{1}{2}(1.2) t^{2}\end{aligned}For t =5 s, you have (Fig. 1)
Motion of B Relative to A. Draw the triangle corresponding to the
vector equation r_{B} = r_{A}+ r_{B/A} (Fig. 2) and obtain the magnitude and direction of the position vector of B relative to A.
Proceeding in a similar fashion (Fig. 2), find the velocity and acceleration of B relative to A. Hence,
REFLECT and THINK: Note that the relative position and velocity of B relative to A change with time; the values given here are only for the moment t = 5 s. Rather than drawing triangles, you could have also used vector algebra. When the vectors are at right angles, as in this problem, drawing vector triangles is usually easiest.
