Question

1. A fish swimming in a horizontal plane has velocity ⃗ where the position relative to a certain rock is ⃗ constant acceleration for 20s, its velocity is ⃗ ( ( ( + ) m/s at a point in the ocean -4 ) m. After the fish swims with -5 ) m/s. a) Find the acceleration of the fish b) If the fish maintains this constant acceleration, where is it at t=25s?

Answer 1

a)

$$a_x=\frac{20-4}{20}=0.8\frac{m}{s^2}\\a_y=\frac{-5-1}{20}=-0.3\frac{m}{s^2}$$

b)

$$x=10+4(25)+0.5(0.8)(25)^2=360\ m\\ y=-4+1(25)-0.5(0.3)(25)^2=72.7\ m$$