A fish swimming in a horizontal plane has velocity V velocity is V-(15.01-3.00 1

Question

A fish swimming in a horizontal plane has velocity V velocity is V-(15.01-3.00 1) m/s. (4.00 i+1.00j) m/s at a point in the ocean where the position relative to a certain rock is (12.0 i 3.00j) m. After the fish swims with constant acceleration for 17.0 s, its (a) What are the components of the acceleration of the fish? ax = D.65 ay-4024 m/s2 m/s2 (b) What is the direction of its acceleration with respect to unit vector i? -19.98counterclockwise from the+-axis (c) If the fish maintains constant acceleration, where is it at t28.0 s? x-378.8V m y=-86.08-. m - In what direction is it moving? -80.10 Draw coordinate axes on a separate piece of paper, and then add the velocity vector with its tail at the origin. Write the numerical values for the x and y components and then use this drawing to determine the angle. counterclockwise from the +x-axis

Explanation / Answer

a) x-component of vector A is Ax = -A*sin(30) = -12*0.5 = -6

b) y-component of vector A is Ay = A*cos(30) = 12*cos(30) =10.4

c) Bx = -B*sin(10) = -8*sin(10) = -1.38

D) By = -B*cos(10) = -8*cos(10) = -7.87

e) Cx = Ax+Bx = -6-1.38= -7.38

Cy = 10.4-7.87 = 2.53

f) Dx = -Cx = 7.38

Dy = -Cy = -2.53

I think you needs only the last one

direction is theta = tan^(-1)(y/x)

theta = tan^(-1)(66.08/378.8) = 9.9 clockwise to the +X-axis

required angle is 360-9.9 = 350.1 degrees counterclockwise to the +X-axis

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