please explain with steps...... where t is in seconds. The magnitude of its inst

Question

please explain with steps......

where t is in seconds. The magnitude of its instantaneous acceleration at time t = 0.5 s is: 20.6 m/s2 20.0 m/s2 14.0 m/s2 12.4 m/s 4.7 m/s2 A particle moves counterclockwise around a circular path of fixed radius around the origin in the (x-y) plane with a constant speed of pi m/s. At time t = 0 it is located at (x = 2, y = 0) m. what is the average velocity of this particle over the time interval t1 = 0 s to t2 = 3 s: 0 m/s pi i^ m/s (pi/3 i^-2 pi/3 j^) m/s -2/3 (i^+j^) m/s -1/3 (i^-j^) m/s

Explanation / Answer

option (D)

initial position =(2i,0)

final position=(0,-2j)

velocity=(displacement )/time=(initial position-final position)/time=(-2j-2i)/3=-(2/3)(i+j)

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