A fish swimming in a horizontal plane has velocity v i = (4.00 i - 5.00 j ) m/s

Question

A fish swimming in a horizontal plane has velocity vi= (4.00 i - 5.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri= (10.0 i - 4.00 j) m. After the fish swims with constant acceleration for 21.0 s, its velocity is v = (23.0 i - 5.00 j) m/s.

(a) What are the components of the acceleration?
ax =  m/s2
ay =  m/s2

(b) What is the direction of the acceleration with respect to unit vector i?
° (counterclockwise from the +x-axis is positive)

(c) If the fish maintains constant acceleration, where is it at t = 30.0 s?
x =  m
y =  m
In what direction is it moving?
° (counterclockwise from the +x-axis is positive)

Explanation / Answer

vi = (4.00 i - 5.00 j) m/s
ri = (10.0 i - 4.00 j) m.
vf = (23.0 i - 5.00 j) m/s.
t = 21 s

(a) What are the components of the acceleration?
ax = [vfx - vix ] / t
    = [23.0 - 4.00] / 21.0
    = 0.905 m/s^2

ay = [vfy - viy ] / t
     = [-5.0 - -5.0] / 21.0
= -0.0 / 21
= 0.0 m/s^2

(b) What is the direction of the acceleration with respect to unit vector i?
° (counterclockwise from the +x-axis is positive)
= tan-1(ay / ax)
   = tan-1(0.0 / 0.905)
   = 0 deg

(c) If the fish maintains constant acceleration, where is it at t = 30.0 s?
at t = 30 s
x = rx + vix t + 0.5 ax t^2
= 10.0 + 4.0 * 30 + 0.5 * 0.905 * 30^2
= 537.25 m

y = ry + viy t + 0.5 ay t^2
= -4.00 + -5 * 30 + 0.5 * 0.0* 30^2
= -154 m

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