A fish swimming in a horizontal plane has velocity Vi-(4.001 After the fish swim

Question

A fish swimming in a horizontal plane has velocity Vi-(4.001 After the fish swims with constant acceleration for 17.0 s, its velocity is v 1 + 1.00 ) m/s at a point in the ocean where the position relative to a certain rock is i,- (12.01-3.80 ) m. (a) What are the components of the acceleration of the fish? ay- 01176m/s? (b) what is the direction of its acceleration with respect to unit vector i? 8.7 Draw coordinate axes on a se parate piece of paper, and then add the acceleration vector components and then use this drawing to determine the angle. counterclockwise from the +x-axis with its tail at th origin, Write the numerical values for the x and y (c) If the fish maintains constant acceleration, where is it at t 28.0 s? -469.74 y 21.89m In what direction is it moving? 357.04x You appear to have given the angle of the velocity after 17.0 seconds, not the velocity after 28.0 seconds counterclockwise from the +x-axis Need Heip?edMaster (33.34 points SerPSE9 4 PD60 My Notes Ask Your T

Explanation / Answer

(b)

direction = tan^-1(ay/ax)

direction = tan^-1(0.1176/0.882) = +352/4

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(c)

velocity after t = 28 s

v2 = v1 + a*t

v2 = (4 i + 1 j ) + (0.882 i - 0.1176 j)* 28

v2 = 28.7 i -2.29 j

direction of motion = tan^-1(vy/vx) = tan^-1(-2.29/28.7) = + 355.4 o

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