A fish is swimming in a horizontal place has velocity of ( vi = 4.00i + 1.00j) ms at a point in the ocean where the position relative to a certain rock is ri=(10.i- 4.00j)m. After the fish swims with constant acceleration for 20 s, it velocity is v = 20.0i -5.00j. Question C) what is the position at 25 s, if at a constant velocity.
Δd= vit+ 1/2at^2
The Attempt at a SolutionOk part a) so the components of V are .79 in the X direction and -0.3 in the Y direction. part b) was you had to find the direction of the vector with respect to the x axis it is 339degree sign. But for part C, I'm solving for Xf and Yf.
Xf = (4.00) (25) + (1/2) (.79)(25)^2 (-- This is wrong. I checked the solution in the back. and the only difference is 10. I think it has to do with the rock. for some reason they randomly added the position of the Rock. So my question is why? Because when I was solving for the components(part A), I didn't need to add the position of the rock. And I got the right answer! Any help would be appreciated