Question

A fish swimming in a horizontal plane has velocity vi = (4.00i + 1.00j ) m/s at a point in the ocean where the position relative to a certain rock is ri = (10.0i - 4.00j) m. After the fish swims with constant acceleration for 20.0 s, its velocity is vf = (20.0i - 5.00j) m/s. (a) What are the components of the acceleration? (b) If the fish maintains constant acceleration, where is it at t =25.0 s?

Answer 1

(a)

$v_{xf}=v_{xi}+a_xt\to a_x=\frac{v_{xf}-v_{xi}}{t}=\frac{20-4}{20}=0.8\ (m/s^2)$

$v_{yf}=v_{yi}+a_yt\to a_y=\frac{v_{yf}-v_{yi}}{t}=\frac{-5-1}{20}=-0.3\ (m/s^2)$

(b)

$\vec r_f=\vec r_i+\vec v_it+\vec a t^2/2=10.0\vec i - 4.00\vec j+(4.00\vec i + 1.00\vec j)\cdot25+(0.8\vec i-0.3\vec j)\cdot25^2/2=$

$=360\vec i-72.75\vec j$.