Question
In: Physics
A particle moves in the xy plane with constant acceleration. At t = 0 the particle...
A particle moves in the xy plane with constant acceleration. At t = 0 the particle is at vector r1 = (3.6 m)i + (2.8 m)j, with velocity vector v1. At t = 3 s, the particle has moved to vector r2 = (11 m)i − (1.8 m)j and its velocity has changed to vector v2 = (4.6 m/s)i − (6.7 m/s)j. (a) Find vector v1. vector v1 = m/s
(b) What is the acceleration of the particle? vector a = m/s2
(c) What is the velocity of the particle as a function of time? vector v(t) = m/s
(d) What is the position vector of the particle as a function of time? vector r(t) = m
Solutions
genius_generous answered 1 year agoRelated Solutions
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