A fish swimming in a horizontal plane has velocity v, = (4.00 î + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is ř, = (16.0 î - 2.40 j) m. After the fish swims with constant acceleration for 19.0 s, its velocity is v = (23.0 î – 6.00 ĵ) m/s. (a) What are the components of the accelerațion of the fish? ax = 1 m/s² -0.37 y m/s² ay (b) What is the direction of its acceleration with respect to unit vector î? 339.40 V ° counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t = 28.0 s? 32 x Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m -9.36 X Y- Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m X = In what direction is it moving? 343.70 V ° counterclockwise from the +x-axis

College Physics
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Chapter1: Units, Trigonometry. And Vectors
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A fish swimming in a horizontal plane has velocity v, = (4.00 î + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is ř, = (16.0 î - 2.40 j) m.
After the fish swims with constant acceleration for 19.0 s, its velocity is V = (23.0 î – 6.00 j) m/s.
(a) What are the components of the acceleration of the fish?
ax = 1 m/s²
m/s2
|-0.37
(b) What is the direction of its acceleration with respect to unit vector î?
339.40 V
° counterclockwise from the +x-axis
(c) If the fish maintains constant acceleration, where is it at t = 28.0 s?
32 x
X =
Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m
-9.36 X
y
Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m
In what direction is it moving?
343.70
° counterclockwise from the +x-axis
Transcribed Image Text:A fish swimming in a horizontal plane has velocity v, = (4.00 î + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is ř, = (16.0 î - 2.40 j) m. After the fish swims with constant acceleration for 19.0 s, its velocity is V = (23.0 î – 6.00 j) m/s. (a) What are the components of the acceleration of the fish? ax = 1 m/s² m/s2 |-0.37 (b) What is the direction of its acceleration with respect to unit vector î? 339.40 V ° counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t = 28.0 s? 32 x X = Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m -9.36 X y Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m In what direction is it moving? 343.70 ° counterclockwise from the +x-axis
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