Answer to Question #217014 in Physics for Dave

Question

A fish swimming in a horizontal plane has velocity Vi=(4i+j)m/s at a point in the ocean where the position relative to a certain rock is ri=(10i -4j)m. After the fish swims with constant acceleration for 20s, its velocity is Vf =(20i – 5j)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

$v_i=4i+j$

$v_f=20i-5j$

$r_i=10i-4j$

(a)

$a=\frac{v_f-v_i}{\Delta t}=\frac{16i-6j}{20}=0.8i-0.3j$

(b)

$r_x=10+4\cdot25+0.8\cdot25^2/2=360\ (m)$

$r_y=-4+1\cdot25-0.3\cdot25^2/2=-72.75\ (m)$

$r_f=360i-72.75j$