Question 12.27: At the instant shown in Fig. 12-45a, cars A and B are travel......

Velocity. The fixed x, y axes are established at an arbitrary point on the ground and the translating x’ , y’ axes are attached to car A, Fig. 12-45a. Why? The relative velocity is determined from v_{B} = v_{A} + v_{B/A} . What are the two unknowns? Using a Cartesian vector analysis, we have

 v_{B} = v_{A} + v_{B/A}

  –  12j = (-18   cos   60°i    –   18  sin  60°j)  +   v_{B/A}

v_{B/A}   = {9i   +   3.588j}   m/s

Thus,

v_{B/A}    = \sqrt{(9)² + (3.588)²} =  9.69 m/s       Ans.

Noting that  v_{B/A}   has +i and +j components, Fig. 12-45b, its direction is

tan θ =  \frac{ (v_{B/A} )_{y}}{ (v_{B/A} )_{x}}  = \frac{3.588}{9}

θ = 21.7°       ⦨         Ans.

Acceleration. Car B has both tangential and normal components of
acceleration. Why? The magnitude of the normal component is

(a_{B})_{n}  = \frac{v²_{B}}{ρ}  = \frac{(12  m/s)²}{100  m} = 1 440 m/s²

Applying the equation for relative acceleration yields

  a_{B} = a_{A} + a_{B/A}

(-1.440i   –   3j) = (2 cos 60°i +  2 sin 60°j) +   a_{B/A}

a_{B/A}  = { -2.440i   –   4.732j}  m/s²

Here  a_{B/A} has -i and -j components. Thus, from Fig. 12-45c,

a_{B/A} = \sqrt{(2.440)² + (4.732)²} = 5.32 m/s²

tan  Φ = \frac{ (a_{B/A} )_{y}}{ (a_{B/A} )_{x}}  = \frac{4.732}{2.440}

Φ  = 62.7°       \measuredangle                Ans.

NOTE: Is it possible to obtain the relative acceleration of a_{B/A} using this method? Refer to the comment made at the end of Example 12.26.