Please answer all empty boxes. And explain step by step, I have another question
ID: 2884278 • Letter: P
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Please answer all empty boxes. And explain step by step, I have another question with different values. A object is attached to S meter long cord and spun in a circle around the point (-4,6). The object's trajectory is given by with tin (m) z (m) Att 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 t 2.5 s? Answer by filing in the vector. (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/s 5.20 m/s Find the formula for s(t) assuming constant velocity stt) v(2.5). Write your answer by filling in both components of the vector below. s (t) Where will the object be 2 seconds after the cord is cut? (C x. Submit Answer Save Progress 02 points Previous Answers Refer to the previous problem. Assume that the cord is cut at 3 seconds instead. Where will the object be 2 seconds later? (C x.Explanation / Answer
When the string is cut, the object moves along hte tangent to the circle at that point.
At 2.5s, x= -2.27, y= 10.69
and slope of r(t) = r'(t) = < -15sin(3t), 15cos(3t) > = <-15sin(7.5), 15cos(7.5)> = <-14.07, 5.2>
S(t) = <-2.27+(-14.07t), 10.69+t*5.2 > = < -2.27-14.07t, 10.69+5.2t >
After 2 seconds after the rod is cut, t=2
S(2) = < -30.41, 21.09 >
b. If the cord is cut at 3 secs, r(3) = < -8.56, 8.06 > ( obtained by substituting t=3 in the equations of r(t) )
Now equation of the tangent at t=3 is
s(t) = <-8.56 +(-14.07t), 8.06+t*5.2 > ( slope of the tangent remains the same )
2 seconds later, S(2) = < -36.7, 18.46 >
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