A fish swimming in a horizontal plane has velocity Vi (4.00 i + 1.00 j) m/\'s at

Question

A fish swimming in a horizontal plane has velocity Vi (4.00 i + 1.00 j) m/'s at a point in he ocean where the position relative to a certain rock is r for 17.0 s, its velocity is v = (23.0 i-1.00 j) m/s. (15.0 1-3.00 j) m. After the fish swims with constant acceleration (a) what are the components of the acceleration of the fish? m/s2 (b) what is the direction of its acceleration with respect to unit vector i? counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t 28.0 ? In what direction is it moving? o counterclockwise from the +xaxis

Explanation / Answer

a)
ax = (v2x - v1x)/t

= (23 - 4)/17

= 1.12 m/s^2

ay = (v2y - v1y)/t

= (-1 - 1)/17

= -0.117 m/s^2

b) direction of a : theta = tan^-1(ay/ax)

= tan^-1(-0.117/1.12)

= 5.96 degrees below +x axis

= 354.04 degrees countecr clockwise from +x axis.

c)

x = xo + vox*t + (1/2)*ax*t^2

= 16 + 4*28 + (1/2)*1.12*28^2

= 567 m

y = yo + voy*t + (1/2)*ay*t^2

= -3 + 1*28 + (1/2)*(-0.117)*28^2

= -20.8 m

d) vfx = vix + ax*t

= 4 + 1.12*28

= 35.4 m/s

vfy = viy + ay*t

= 1 + (-0.117)*28

= -2.28 m/s

direction : theta = tan^-1(vfy/vfx)

= tan^-1(-2.28/35.4)

= 356.3 degrees counetcrclockwise from +x axis

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