A car is travelling around a horizontal circular track with radius r = 240. 0 m

Question

A car is travelling around a horizontal circular track with radius r = 240. 0 m as shown. It takes the car t = 54. 0 s to go around the track once. the angle theta A = 20. 0 degree above the x axis, and the angle theta 5 = 61. 0 degree below the x axis. What is the magnitude of the car's acceleration? m/s2 What is the x component of the car's acceleration when it is at point A m/s2 What is the y component of the car's acceleration when it is at point A m/s2 What is the x component of the car's acceleration when it is at point B m/s2 What is the y component of the car's acceleration when it is at point B m/s2 As the car passes point B, the y component of its acceleration is increasing constant decreasing

Explanation / Answer

(a) omega(angular speed) = 2 x pi /54 , alpha=angular acceleration = r x omega = 27.92 rad/s2 , a = r x alpha = 6700m/s2 Ans. (b) x component of car's velocity at A = vsin(20) = r x omega sin(20) = r x (2 x pi/54) sin(20) = 9.55 (-i)m/s Ans. (c) y component of car's velocity at A = vcos(20) = r x omega cos(20) = r x (2 x pi/54) cos(20) = 26.24 (j)m/s Ans. (d) x component of car's acceleration at B = asin(61) = 5860(-i)m/s2 Ans. (e) y component of car's acceleration at B = acos(61) = 3248.22(j)m/s2 Ans. (f) theta decreases so cos (theta) increases so y component of velocity keeps increasing.

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