The position of a particle moving in an XY direction is defined by the vector function r(t) = {(2)t^3 - (5)t}i + {(6) - (7t^4)}j. For t = 2 seconds, the position r(2) is calculated as r(2) = (2(2)^3 - 5(2))i + (6 - 7(2)^4)j, resulting in r(2) = 8i - 110j. The velocity v(t) is derived as v(t) = dr/dt, yielding v(2) = (12)t^2 - 5i + (-28)t^3j, which evaluates to v(2) = 44i - 224j. The acceleration a(t) is found by differentiating the velocity, resulting in a(t) = d^2r/dt^2, leading to a(2) = 24ti - 84t^2j, which gives a(2) = 48i - 336j.
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Welcome to Physics Forums.Saba Gaad said:The Position of a particle moving in an XY direction is given by r={(2)t^3-(5)t }i + { (6)-(7t^4) }j. Calculate
(a) r
(b) v
(c) a
when t=2 sec.