A fish swimming in a horizontal plane has velocity V_i vector = (4.00 i + 1.00 j

Question

A fish swimming in a horizontal plane has velocity V_i vector = (4.00 i + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is r_i vector = (16.0 i - 1.60 j) m. After the fish swims with constant acceleration for 17.0 s, its velocity is V = (19.0 i - 6.00 j) m/s. What are the components of the acceleration of the fish? What is the direction of its acceleration with respect to unit vector i? counterclockwise from the +x-axis If the fish maintains constant acceleration, where is it at t = 26.0 s? In what direction is it moving? counterclockwise from the +x-axis

Explanation / Answer

a)

a = (v-vi)/t = [(19.0 i - 6.0 j) - (4.00 i + 1.00 j)]/17 = [15i - 5j]/17
ax = 15.0/17.0 m/s2

=0.64706 m/s^2
ay = -5/17.0 m/s2

= -0.2353m/s^2

b)
tan^-1(-5/15) = -18.43°
-18.43° + 360°

= 341.56°

c)

r = ri + (vi)t + (1/2)at^2

= (16.0 i - 3.2 j) + 26(4.00 i + 1.00 j) + (26.0^2)(1/2)[0.647i - 0.235j]

= (120.00 i + 22.80 j) + [218.6i - 79.43j]

= (338.6i - 56.63j) m

x = 338.6 m
y = - 56.63 m

v = vi + at = (4.00 i + 1.00 j) + 26[0.647i - 0.235j]

= (20.822i - 5.11j) m/s

tan^-1(-5.11/20.822) = -14.5°

-13.78° + 360° =346.2°

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