Answer to Question #258121 in Mechanics | Relativity for isa

Question

At the instant shown, cars A and B are traveling at speeds of 30 km/h and 20 km/h, respectively. If A is increasing its speed at 400 km/h² whereas the speed of B is decreasing at 800 km/h², determine the velocity and acceleration of B with respect to.

Answer 1

Normal acceleration of B:

$(a_B)_n = \frac{v_B^2}{p} = \frac{20^2}{0.3}=1333.34 \;km/h^2 \\ Normal\; acceleration\; vector \; of \; B: \\ (a_B)_n = (a_b)_n[cos 30°i -sin 30°j] \\ = 1333.34[cos 30°i -sin 30°j] \\ = 1154.67i -666.67j$

Tangential acceleration vector of B

$(a_B)_t = (a_B)_t [-sin30°i -cos30°j] \\ = 800[-sin30°i -cos30°j] \\ = -400i -692.8j \\ Total \;acceleration \;of \;B \\ a_B = (a_B)_t +(a_B)_n \\ = -400i -692.8j +1154.67i -666.67j \\ = 754.67i -1359.47j \\ Acceleration \; of\; B\; with \; respect\; to \; A. \\ a_{B/A} = a_B -a_A \\ = (754.67i -1359.47j) -400j \\ = 754.67i -1759.47j \\ a_{B/A} = \sqrt{(754.67)^2 +(1759.47)^2} \\ = 1914.487 \;km/h^2 \\ θ = tan^{-1} (\frac{-1759.47}{754.67}) \\ θ = -66.784°$

Velocity

$v_B = -v_bcos 30°i+v_Bsin30°j \\ = -20cos30°i +20sin30°j \\ v_B = -17.32i +10j \\ v_A = -v_Ai =-30i \\ v_B =v_A +v_{B/A} \\ v_{B/A} = v_B -v_A \\ = -17.32i +10j +30i \\ = 12.68i +10j \\ = \sqrt{(12.68)^2+(10)^2} \\ = 16.148 \;km/h \\ θ = tan^{-1}(\frac{10}{12.68}) \\ θ = 38.26°$