A fish swimming in a horizontal plane has velocity vi = (4.00 i - 1.00 j) m/s at

Question

A fish swimming in a horizontal plane has velocity vi = (4.00 i - 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri = (-10.0 i + 4.00j) m. After the fish swims with constant acceleration for 23.0 s, its velocity is v = (23.0 i - 5.00 j) m/s.
(a) What are the components of the acceleration?
ax=?
ay=?
(b) What is the direction of the acceleration with respect to unit vector i?
(c) If the fish maintains constant acceleration, where is it at t = 28.0 s?
x=?
y=?
In what direction is it moving?
_____° (counterclockwise from the +x-axis is positive)

Explanation / Answer

Here ,

initial velocity , vi = 4 i - j m/s

initial position , ri = -10 i + 4 j m

time , t = 23 s

final velocity , vf = 23 i - 5 j m/s

a)

acceleration = (vf - vi)/time

acceleration = ((23 i - 5 j ) - ( 4 i - j ))/23

acceleration = 0.83 i - 0.174 j m/s^2

the components are :

ax = 0.83 m/s^2

ay = -0.174 m/s^2

b)

for the direction of acceleration = arctan(ay/ax)

direction of acceleration = arctan(-0.174/.83)

direction of acceleration = -11.8 degree C

the direction of acceleration is 11.8 degree below i

c)

for the final position after 28 s

r = ri + u * t + 0.5 * at^2

r = -10 i + 4 j + (4 i - j) * 28 + 0.5 * 28^2 *(0.83 i - 0.174 j)

r = 427.4 i - 92.21 j m

the position of the fish is

x = 427.4 m , y = -92.21 m

------------------

for the direction of motion

vf = vi + a * t

vf = 4 i - j + 28 * (0.83 i - 0.174 j)

vf = 27.24 i - 5.872 j m/s

direction of motion = arctan(-5.872/27.24)

direction of motion = -12.2 degree

the direction of motion is -12.2 degree

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