A fish swimming in a horizontal plane has velocity vi = (4.00 i + 4.00 j) m/s at

Question

A fish swimming in a horizontal plane has velocity vi = (4.00 i + 4.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri = (-10.0 i + 4.00 j) m. After the fish swims with constant acceleration for 25.0 s, its velocity is v = (19.0 i - 5.00 j) m/s.

(a) What are the components of the acceleration?
(b) What is the direction of the acceleration with respect to unit vector i?
(c) If the fish maintains constant acceleration, where is it at t = 35.0 s?
In what direction is it moving?

Explanation / Answer

19=4+25a

ai=.6

aj=-5-4/25=-.36

a=.6i-.36j

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