Question

2. -12 points PSE6 4.P.007 My Notes A fish swimming in a horizontal plane has velocity vi = (4.00 İ + 5.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri = (-10.01-4.00 j) m. After the fish swims with constant acceleration for 24.0 s, its velocity is v-(21.0-5.00) m/s. (a) What are the components of the acceleration? ax = m/s2 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-34.0 s? In what direction is it moving? (counterclockwise from the +x-axis is positive)If the number is a decimal please include 4 digits past the decimal point if available.

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Answer #1

(a) acceleration = change in velocity / time interval

acceleration = (21i - 5j) - (4i + 5j) / 24

acceleration = 17i - 10j / 24

acceleration = 0.70833i - 0.4166j

Therefore,

ax = 0.70833

ay = - 0.4166

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(b) direction = arctan (-0.4166 / 0.70833)

direction = 30.46 degree below the +x axis or 329.5 degree counterclockwise from positive x axis.

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(c) x = -10 + 4*34 + 0.5*0.70833 * 342

x = -10 + 136 + 409.41

x = 535.41 m

y = - 4 + 5*34 - 0.5*0.4166 *342

y = -4 + 170 - 240.8

y = -74.8 m

direction will be

theta = arctan (5 - 0.4166 * 34 / 4 + 0.70833 *34)

theta = arctan ( -9.1644 / 28.1)

theta = 18.1 degree ( or 341.9 degree counterclockwise from + x axis)

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