A bicycle, initially located xo = 2.3 m at time t-D0 seconds, has a velocity at time t given by v = 1.6 t3 + 5.6t with v in m/s and t in seconds. What is the position x (in meters) and the acceleration a (in m/s2) of the bicycle at time t? O x = 0.53 t + 5.6 t2 + 2.3, a = 4.8 t2 + 5.6 O x = 0.4 t + 2.8 t2 + 2.3 a = 4.8 t2 + 2.3 Ox = 4.8 t² + 5.6 a = 0.4 t + 2.8 t² + 2.3 O x = 4.8 t2 +2.3 a = 0.4 t + 2.8 t2 + 2.3 Ox= 0.4 t + 2.8 t2 + 2.3, a= 4.8 t2 + 5.6

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Chapter1: Units, Trigonometry. And Vectors
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A bicycle, initially located xo = 2.3 m at time t=0 seconds, has a velocity at time t given by
%3D
v = 1.6 t3 + 5.6 t
with v in m/s and t in seconds. What is the position x (in meters) and the acceleration a (in m/s-) of
the bicycle at time t?
O x= 0.53 t + 5.6 t2 + 2.3
a = 4.8 t2 + 5.6
O x = 0.4 t + 2.8 t2 + 2.3
a = 4.8 t2 + 2.3
O x = 4.8 t2 + 5.6
a = 0.4 t + 2.8t + 2.3
O x= 4.8 t2 + 2.3
a = 0.4 t + 2.8 t2 + 2.3
Ox = 0.4 t + 2.8 t2 + 2.3
a = 4,8 t2 + 5.6
Transcribed Image Text:A bicycle, initially located xo = 2.3 m at time t=0 seconds, has a velocity at time t given by %3D v = 1.6 t3 + 5.6 t with v in m/s and t in seconds. What is the position x (in meters) and the acceleration a (in m/s-) of the bicycle at time t? O x= 0.53 t + 5.6 t2 + 2.3 a = 4.8 t2 + 5.6 O x = 0.4 t + 2.8 t2 + 2.3 a = 4.8 t2 + 2.3 O x = 4.8 t2 + 5.6 a = 0.4 t + 2.8t + 2.3 O x= 4.8 t2 + 2.3 a = 0.4 t + 2.8 t2 + 2.3 Ox = 0.4 t + 2.8 t2 + 2.3 a = 4,8 t2 + 5.6
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