Question

A particle of mass 0.50 kg starts moves through a circular path in the xy-plane with a position given by →r(t)=(4.0cos3t)i+(4.0sin3t)j where r is in meters and t is in seconds.

Find the velocity and acceleration vectors as functions of time.
a) v(t)=−12.0sin(3t) i +12.0cos(3t) j
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b) v(t)=−4.0sin(3t) i+4.0cos(3t)j
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c) v(t)=−12.0cos(3t) i+12.0sin(3t)j
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d) v(t)=4.0cos(3t) i−4.0sin(3t)j

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Answer 1

The velocity vector is v(t) = -12.0sin(3t) i + 12.0cos(3t) j. The acceleration vector is a(t) = -36.0cos(3t) i - 36.0sin(3t) j.

Step-by-step explanation:

The particle's position vector is given by r(t) = (4.0 cos 3t)i + (4.0 sin 3t)j. To find the velocity vector, we can differentiate the position vector with respect to time. The velocity vector is given by v(t) = -12.0sin(3t) i + 12.0cos(3t) j.

To find the acceleration vector, we can differentiate the velocity vector with respect to time. The acceleration vector is given by a(t) = -36.0cos(3t) i - 36.0sin(3t) j.