A fish swimming in a horizontal plane has velocity vi = (4.00 i + 3.00 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 23.0 s, its velocity is v = (17.0 i - 5.00 j) m/s.
(a) What are the components of the acceleration?
ax =
.57
Correct: Your answer is correct.
m/s2
ay =
-.35
Correct: Your answer is correct.
m/s2
(b) What is the direction of the acceleration with respect to unit vector i?
-31.55
Correct: Your answer is correct.
° (counterclockwise from the +x-axis is positive)
(c) If the fish maintains constant acceleration, where is it at t = 29.0 s?
x =
365.69
Correct: Your answer is correct.
m
y =
-64.175
Incorrect: Your answer is incorrect.
Your response differs from the correct answer by more than 10%. Double check your calculations. m
In what direction is it moving?
-
Incorrect: Your answer is incorrect.
° (counterclockwise from the
Porcia C.
10/16/16