A fish swimming in a horizontal plane has velocity = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to a certain rock is = (14.01 - 2.60 ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (23.01 - 1.00 ) m/s. (a) What are the components of the acceleration of the fish? ax = 1.27 m/s m/s ay (b) What is the direction of its acceleration with respect to unit vector ? counterclockwise from the +x-axis -5.98 = -0.133 (c) If the fish maintains constant acceleration, where is it at t = 28.0 s? 476 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 16.9 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 In what direction is it moving? Need Help? Submit Answer o counterclockwise from the +x-axis Read It Master It