A fish swimming in a horizontal plane has velocity = (4.00 + 1.00 ) m/s at a poi

Question

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 = (12.0 - 2.80 ) m. After the fish swims with constant acceleration for 19.0 s, its velocity is = (23.0 - 1.00 ) m/s. What are the components of the acceleration of the fish? What is the direction of its acceleration with respect to unit vector ? counterclockwise from the +x-axis If the fish maintains constant acceleration, where is it at t = 30.0 s? In what direction is it moving? counterclockwise from the +x-axis

Explanation / Answer

a) ax = (vxf - vxi)/t = (23-4)/19=19/19=1 m/s^2

ay = (vyf- vyi)/t =(-1-1)/19=-2/19=-.105 m/s^2

b) =arctan(ay/ax)=arctan(-.105)=-5.994 degrees

c) x = x0 + v0x t + 1/2 a t^2

x = 12 + 4*30+1/2*1*30^2 = 582

y = y0 + v0y t + 1/2 a t^2= -2.8 + 1*30+1/2*(-.105)*30^2=-20.05

to get direction that its moving we need velocities

vx = v0x + at = 4+1*30 = 34

vy = 1 + -.105*30=-2.15

direction = arctan(vy/vx)= arctan(-2.15/34)=-3.62 degrees

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