Question

Refer to the previous problem. Assume that the cord is cut at 3 seconds instead. Where will the object be 2 seconds later? 36.7 m X 18.46 m Previous Problem A object is attached to 5 meter long cord and spun in a circle around the point (-4,6). The object's trajectory is given by r (t)-(5 cos(3t) 4, 5sin(3t) 6) m with t in seconds y (m) 12 r (m) 10 8 42 -2 At t 2.5 seconds the cord is cut. From that instant onward, the object moves in a straight line, s (t), in the direction of its tangent vector. Answer the following questions. All decimals must be accurate to at least two places. (Maybe more, since you don't want to accumulate roundoff error.) What is the position at that instant t2.5 s? Answer by filling in the vector. s (2.5)2.27 m 10.69 m What is the velocity the instant t 2.5 s? Answer by filling in the vector v (2.5)14.07 m's5.2 ms Find the formula for s(t) assuming constant velocity s (t)v (2.5) Write your answer by filling in both components of the vector below 32.905-14.07 -2.315.2t Where will the object be 2 seconds afte the cord is cut? 30.41 m 21.09 m

Answer 1

Given data A object is attached to 5 meter long cord and span in a circle around in the point (-4, 6). The object's trajectory is given by r^vector (t) = (5 cos (3t) - 4, 5 sin (3t) + 6)m with t in seconds. At t = 2.5 seconds The cord is cut. From mat instant onward, The object moves in a straight joint line, s^vector (t), in me direction of its tangent vector. \r\n s^vector (2, 5) = < -2.2 cm. 10.69 m what is me velocity me instant t = 2.55 V^vector (2, 5) = < -14.07 m/s 5.2 m/s me Formula for s^vector (t) constant velocity s^vector (t) = V^vector (2.5) filling in both components of me vector. s^vector (t) = < 32.905 - 14.07 t -2.31 + 5.2 t The object be 2 seconds after me card is cut. < -30.41 m. 21.09 m given mat The object's trajectory r^vector (t) = (5 cos t, 5 sin t)m Eliminating me parameter