A fish swimming in a horizontal plane has velocity v with arrowi = (4.00 i + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is r with arrowi = (12.0 i − 2.00 j) m. after the fish swims with constant acceleration for 17.0 s, its velocity is v with arrow = (23.0 i − 1.00 j) m/s. (a) what are the components of the acceleration of the fish? ax = m/s2

Respuesta :

To determine the acceleration of the fish, we use one of the kinematic equation which relates velocity and acceleration. Using the given velocities, we substitute it to the equation. As we see, the velocities include their directions where i represents the x direction and j represents the y direction. We do as follows:

a = (v-vi) / t
a = [ (20.0 i - 5.00 j) - (4.00 i + 1.00 j)] / 23
Combining like terms,
a = [16i - 6j] / 23
TO determine the components of the acceleration, we do as follows:
 
ax = 16.0/23.0 m/s2 = 0.70 m/s^2 ( to the right)
ay = -6.0/23.0 m/s2  = -0.26 m/s^2 ( accelerating down)