Question

Car B turns such that its speed is increased by (Q.)=(0.5e)m/s, where t is in seconds. If the car starts from rest when =0°, determine the magnitude of its velocity and acceleration when the arm AB rotates =30°. Neglect the size of the car. (Note: using tangent and normal components to solve this problem)

Answer 1

Given data: Tangential acceleration (at) = (0.5et) m/sec? 0 = 0 deg, 0 = 30 deg = 30* (T / 180) = 0.52359 radian. Distance traveled in 30 deg (s) = r * = 5 * 0.52359 2.618 m. By definition we know that: Acceleration (a) = dv/dt dv = a .dt After integration on both sides, s dv = fa.dt of dv = of* (0.5e). Dt v=0.5 * (e - 1) By definition we know that: Velocity (v) = ds / dt v. dt = ds After integration on both sides, ost (0.5*(e - 1)) = 0[2.618 ds 0.5 * (et - t - 1) = 2.618 After solving this we get, time (t) = 2.123 seconds. Magnitude of its velocity when arm rotates 0 = 30 deg: v = 0.5*(et - 1) = 0.5 * (22.123 - 1) = 3.68 m/sec.... Answer.

Tangential acceleration (at) = (0.5et) = 4.18 m/sec? Normal acceleration (an) = v2 / r = 3.68^2/5 = 2.7085 m/sec. Magnitude of its acceleration when arm rotates 0 = 30 deg: a = sqrt (an?) + (ar?)) = sqrt((2.7085^2)+(4.1842)) = 4.9808 m/sec.... Answer.