fish swimming in a horizontal plane has velocity vi 4.00 1.00 j) m/s at a paint
ID: 1580096 • Letter: F
Question
fish swimming in a horizontal plane has velocity vi 4.00 1.00 j) m/s at a paint in the ocean where the position relative to a certain rock is r = 16.0 i-2.20 j m Atter the tish swims with constant acceleration tor 15.0 s, its velocity s V-(17.0 1-3.00 j) m/s (a) what are the components of the acceleration of the fish? m/s m/s2 (b) What is the direction of its acceleration with respect to unit vector i? e counterclockwise from the +x-axls (c) If the fish maintains constant acceleration, where is it at t 26.0 s? In what direction is it moving? e counterclockwise from the x-axlsExplanation / Answer
(a)
acceleration a = (v2 - v1)/t1
acceleration a = ((17i - 3j ) - (4i + 1j))/15
acceleration a = 0.87 i - 0.27 j
ax = 0.87 m/s^2 <<<<<------------ANSWER
ay = -0.27 m/s^2 <<<<<------------ANSWER
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(b)
direction = tan^-1(ay/ax) = 343 counterclockwise from + axis <<<<<------------ANSWER
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(c)
r2 - r1 = v1*t + (1/2)*a*t^2
r2 -( 16i - 2.2j ) = (4i +j)*26 + (1/2)*(0.87i - 0.27j)*26^2
r2 - 16i + 2.2j = 104i + 26j + 294.1i - 91.26 j
r2 = 414.1 i - 67.46 j
x = 414.1 m <<<<<------------ANSWER
y = -67.46 m <<<<<------------ANSWER
velocity after t = 26 s
v = v1 + a*t
v = 4i + j + (0.87i - 0.27j)*26
v = 26.62i - 6.02 j
direction of motion = tan^-1(-6.02/26.62) = 167.3 <<<<<------------ANSWER
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