Curvilinear Motion: Rectangular Components Learning Goal: To be able to calculat

Question

Curvilinear Motion: Rectangular Components Learning Goal: To be able to calculate position, velocity and acceleration of an object in curvilinear motion using a rectangular coordinate system. An object's motion can be described along a path represented by a fixed x, y, z coordinate system. In such a system the position vector, r, is described as r = xi + yj + zk The object's velocity, v, can be found by taking the first time derivative of the position vector, r. v = dx/dt = v_x i + v_y j + v_z k The object's acceleration a, can be found by taking the time derivative of the velocity, v. a = dv/dt = a_x i + a_y j + a_z k Part A A car drives on a curved road that goes down a hill. The car's position is defined try the position vector. r = {- [30.0 cos(pi/10.0 t)]i + [30.0 sin(pi/10.0 t)]j - (A_zt)k} ft where A_z = 25.0ft/s. The image below shows the system projected onto the x-y plane. What are the car's velocity and acceleration vectors at this position. Draw your vectors starting at the black dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded Part B What ¡s the magnitude, v, of the car's velocity, v, at t = 3 00 s? Express your answer numerically in feet per second to three significant figures. v = ___________________ ft/s

Explanation / Answer

r = (-(30cos(pi*t/10))i +(30sin(pi*t/10))j - (Azt)k)ft

Az = 25 ft/s

r = (-(30cos(pi*t/10))i +(30sin(pi*t/10))j - (25t)k)ft

v = dr/dt

v = (((30*pi/10)sin(pi*t/10))i +((30*pi/10)cos(pi*t/10))j - (25)k)ft

v = (((30*pi/10)sin(pi*3/10))i +((30*pi/10)cos(pi*t/10))j - (25)k)ft

v = 7.625 i + 5.54 j - 25 k

magnitude v = (7.625^2 +5.54^2 +25^2)^0.5

v = 26.7 ft/s

a = (((30*pi^2/100)cos(pi*t/10))i -((30*pi^2/100)sin(pi*t/10))j

a = (((30*pi^2/100)cos(pi*3/10))i -((30*pi^2/100)sin(pi*3/10))j

a = 1.74 i - 2.4 j

a = (1.74^2 +2.4^2)^0.5

a = 2.96 m/s^2

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