Please answer the question ASAPthank you **Problem 2: Wheel and Pulley System Analysis****Objective:**Given the information provided at an instant, determine the angular velocity and angular acceleration of the wheel C.**Problem 2 Statement:**At the given instant, point B has a velocity of 4 ft/sec and an acceleration of 2 ft/sec² downwards. Determine the acceleration and the angular acceleration of point A.**Diagram Explanation:**- The diagram illustrates a pulley and wheel system.- Wheel C, located at the top of the system, has a radius of 150 mm.- There is a pulley on the left that is connected to a wheel at point A.- The distance from point A to the point of contact below the pulley is 400 mm.- The velocity of B is indicated as \( v_B = 4 \, \text{m/s} \) downwards.- The acceleration of B is indicated as \( a_B = 2 \, \text{m/s}^2 \) downwards. This setup likely involves rotational motion principles and requires conversion between linear motion of point B and angular motion of the wheel.

Given, Vb = 4m/s Ab = 2 m/s2Using vector analysis Velocity equation:- Vb = Va + Wab x Yba Where Va = -Vaj , Wab = Wab k, Yba = 0.3i – 0.4j m and Vb = 4i m/s 4i = -Vaj + [.i j k] [ 0 0 Wab] [ 0.3 -0.4 0] 4i = -Vaj + i(0.4Wab) – j(0-0.3Wab) + k(0) 4i = -Vaj + 0.4Wabi + 0.3Wabj On comparing coefficients of I and j, 4 = 0.4 Wab Wab = 4/04 = 10 rad/s And -Va + 0.3Wab = 0 Va = 0.3 x 10 = 3 m/s Now, Acceleration eqn Ab = Aa + αab x Yba – W2ab . Yba Where Ab = 2i , Aa = (V2a / Y)I – αac(0.15)j Aa = 60i – 0.15 αacj m/s2 Αab = αabk , and Yba = 0.3i – 0.4 j m 2i = 60i – 0.15αacj + [.i j k] – 102(0.3i – 0.4j) [0 0 αab] [0.3 -0.4 0] 2i = 60i – 0.15αacj + i(0.4αab) – j(-0.3αab) – 30i + 40j On comparing coefficient, 2 = 60 + 0.4αab – 30 .αab = -70 rad/s2 And 0 = -15αac + 0.3αab + 40 . αac = (0.3)(-70) + 40 / 0.15 . αac = 126.6 rad/s And Wac = Va/Y = 3/0.15 = 2.0 rad/s