Question

152. A car is traveling on a spiral section of a road. The spiral path of the road is described by the cquation r 3800 meters with e in radians. The angular velocity of the car is a constant 0.040 radis i direction shown. Find the radial and transverse components of velocity and accelerat when θ-60°. Show these two components of velocity and acceleration on the sketches the components of velocity and acceleration pointing in their proper directions In addition, find the magnitude of velocity and acceleration when 9-6 below. Shonw Origin Origin Velocity Acceleration 22: Radial Acceleration Transverse Acceleration Radial Velocity Transverse Velocity Magnitude of Acceleration Magnitude of Velocity Show the com s of velocity acceleration on the sketches above aNSweYS S. I5.92 O. 637 2 22.0 137 ar de

Answer 1

Therefore r = 380 theta and d theta/dt = theta degree = 0.040 rad/s (given) dr/dt = 380 d theta/dt r degree = 380 theta degree = 380 times 0.040 r degree = 15.2 m/s d^2r/dt^2 = r degree = 0 angular velocity is constant so, theta degree = 0 when theta = 60 degree = product/3 rad V_r = r degree = 15.2 m/s (radial velocity) r_theta = r theta degree = 15.2 times 0.040 = 0.608 m/s v_theta = 0.608 m/s (transverse velocity) v = squareroot v_r^2 + v theta^2 = squareroot (15.2)^2 + (0.608)^2 v = 15.21 m/s (magnitude of net velocity) Now for acceleration a_r = r degree - r (theta degree)^2 = 0.380 theta(theta degree)^2 a_r = -380 (pi/3) (0.040)^2 = -0.636 m/s a_r = -0.636 m/s a_theta = r theta^.. + 2r^. theta^. = 0 + 2(15.2) (0.04) a_theta = 1.216 m/s magnitude of acceleration, a = squareroot a_theta^2 + a_r^2 = squareroot (1.216)^2 + (- 0.636)^2 a = 1.37 m/s^2