A 5.00-kg particle starts from the origin at time zero. Its velocityasa functionoftimeisgivenbyv = (6t 2i+ 2t j) where v is in meters per second and t is in seconds. (a) Find its position vector r as a function of time (b) its acceleration as a function of time,(c) the net force exerted on the particle as a function of time,
A 5.00-kg particle starts from the origin at time zero. Its velocityasa functionoftimeisgivenbyv = (6t 2i+ 2t j) where v is in meters per second and t is in seconds. (a) Find its position vector r as a function of time (b) its acceleration as a function of time,(c) the net force exerted on the particle as a function of time,
An Introduction to Physical Science
14th Edition
ISBN:9781305079137
Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Chapter2: Motion
Section2.3: Acceleration
Problem 2.3CE: What is the average speed in mi/h of a person at the equator as a result of the Earths rotation?...
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- A 5.00-kg particle starts from the origin at time zero. Its velocityasa functionoftimeisgivenbyv = (6t 2i+ 2t j) where v is in meters per second and t is in seconds. (a) Find its position vector r as a function of time (b) its acceleration as a function of time,(c) the net force exerted on the particle as a function of time, (d) the net torque about the origin exerted on the particle as a function of time,(e)the
angular momentum of the particle as a function of time,(f)the kinetic energy of the particle as a function of time, and (h)the power injected into the system of the particle as a function of time.
Hint: (a) v = dr/dt and dr = vdt, r =òvdt (integrate v and get r),(b) a=dv/dt, © F=ma,(d) t = r x F (cross product) (e) L= r x mv(cross product) (f) KE=(1/2)mv2(Square v vector to get v2,i2 =j2 =1),(g) Power = first derivative of KE
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