A fish swimming in a horizontal plane has velocity V_i vector = (4.00 i cap + 1.

Question

A fish swimming in a horizontal plane has velocity V_i vector = (4.00 i cap + 1.00 j cap) m/s at a point in the ocean where the position relative to a certain rock is r_i vector = (16.0 i cap - 3.20 j cap) m. After the fish swims with constant acceleration for 17.0 s, its velocity is v vector = (15.0 i cap - 3.00 j cap) 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 cap? If the fish maintains constant acceleration, where is it at t = 26.0 s? In what directions is it moving?

Explanation / Answer

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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