Question

a fish swimming in a horizontal plane has a velocity v(i)=(4.00i+1.00j) m/s at a point in the ocean where the position realative to a certain rock is r(i)=(10.0i-4.00j)m. After the fish swims with constant acceleration for 20.0 s, its velocity is v=(20.0i-5.00j)m/s. a) what are the components of the acceleration of the fish? what is the direction of its acceleration with respect to unit vector (i) what if the fish maintains constant acceleration, where is it at 25.0 sec and in what direction is it moving.

Answer 1

As the acceleration is constant, decompose the motion into x and y components
u=4
v=20
t=20
and
u=1
v=-5
t=20

Use a=(v-u)/t to find a

It is
a=0.8
and
a=-0.25

S0\[v(t)=(4-0.8t)i,(1-0.25t)j\]Integrating:\[r(t)=(4t-0.4t^2)i,(1t-0.125t^2)j\]

Answer 2

Sorry, that was minus 4, not 5, a is still -0.25

Answer 3

I'm not sure if that unit vector refers to the fish's position or velocity.

Answer 4

For the 2nd question, just plug in 25 to v(t) and r(t)