A) A particle rotates counterclockwise in a circle of radius 6.7 m with a consta

Question

A) A particle rotates counterclockwise in a circle
of radius 6.7 m with a constant angular speed
of 13 rad/s . At t = 0, the particle has an x
coordinate of 5.1 m and y > 0 .
Determine the x coordinate of the particle
at t = 1.21 s .
Answer in units of m
B) Find the x component of the particle’s veloc-
ity at t = 1.21 s.
Answer in units of m/s
C) Find the x component of the particle’s accel-
eration at t = 1.21 s.
Answer in units of m/s2

Explanation / Answer

Assuming the center of the circle is at (0,0), then y = sqrt((6.7m)² - (5.1m)²) = 4.345 m If x(t) = Acos(?t - f), then A = 6.7m and x(0) = 5.1m = 6.7m * cos(?*0 - f), which leads to f = -0.7056 rad (make sure your calculator is in radians!) ? = 13 rad/s (given), so Position --> x(t) = 6.7*cos(13t + 0.7056) Velocity --> v(t) = -A?*sin(?t - f) = -6.7*13*sin(13t + 0.7056)= -87.1*sin(13t + 0.7056) Acceleration --> a(t) = -A?²*cos(?t - f) = -6.7*13²*cos(13t + 0.7056) = -1132.3*cos(13t + 0.7056) at t = 1.21s, T = 13*1.21 + 0.7056 = 16.4356 Velocity: v(1.21) = -87.1*sin(13*1.21 + 0.7056) = 57.93 m/s The x-component of the velocity is Vx = v*cosT = 57.93*cos 16.4356 = -43.26 m/s Acceleration: a(1.21) =-1132.3*cos(13*1.21 + 0.7056)) = 845.54 m/s² The x-component of the acceleration is Ax = a*cosT = -631.41 m/s²

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