A fish swimming in a horizontal plane has velocity v = ((4.00)î + (1.00)f)m/s at a point in the ocean where its position relative to a certain rock is r = ((10.0)î – (4.00)f)m. After the fish swims with constant acceleration for 20.0 s, its velocity becomes v ((20.0)-(5.00)f)m/s. What are the horizontal and vertical components of the acceleration (in m/s2)? %3D O a. -1.4, 2.9 ОБ.1.4, 2.6 Ос. -0.3, 0.8 Od.0.8, -0.3 De. 2.4, -1.5

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Chapter1: Units, Trigonometry. And Vectors
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A fish swimming in a horizontal plane has velocity vị = ((4.00)î + (1.00)ĵ)m/s at a point in the ocean where its position
relative to a certain rock is T= ((10.0)î –- (4.00)j)m. After the fish swims with constant acceleration for 20.0 s, its
velocity becomes v = ((20.0)î – (5.00)§)m/s. What are the horizontal and vertical components of the acceleration (in
m/s2)?
%3D
%3D
O a.-1.4, 2.9
ОБ.1.4, 2.6
Oc. -0.3, 0.8
Od.0.8, -0.3
O e. 2.4, -1.5
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Transcribed Image Text:uestion Completion Status: A fish swimming in a horizontal plane has velocity vị = ((4.00)î + (1.00)ĵ)m/s at a point in the ocean where its position relative to a certain rock is T= ((10.0)î –- (4.00)j)m. After the fish swims with constant acceleration for 20.0 s, its velocity becomes v = ((20.0)î – (5.00)§)m/s. What are the horizontal and vertical components of the acceleration (in m/s2)? %3D %3D O a.-1.4, 2.9 ОБ.1.4, 2.6 Oc. -0.3, 0.8 Od.0.8, -0.3 O e. 2.4, -1.5 Question 3 of A Moving to the next question prevents changes to this answer. Close Windo
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