Question

A fish swimming in a horizontal plane has velocity V (4.00 i1.00 j) m/s at a point in the ocean where the position relative to a certain rock is fter the fish swims with constant acceleration for 19.0 s, its velocity is V (15.0 1-3.00 j) m/s (16.0 1 4.20) m (a) What are the components of the acceleration of the fish? ax0.579 m/s2 0.105X Review the definition of average acceleration and remember that each component is treated separately. m/s2 a, (b) What is the direction of its acceleration with respect to unit vector i? 349.7 X You appear to have correctly calculated the angle using your incorrect values from part (a).° counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t 28.0 s? x-354.968m 73.36 X The time given in part (c) is the total time the fish has been swimming. Do not add it to the time from part (a). m In what direction is it moving? 48.97 X Draw coordinate axes on a separate piece of paper, and then add the velocity vector with its tail at the origin. Write the numerical values for the x and y components and then use this drawing to determine the angle.° counterclockwise from the +x-axisHelp!

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Answer #1

Given,

vi = 4 i + 1 j ; t = 19 s ; v = 15 i - 3 j

a)we know that

a = (v2 - v1)/t = (v - vi)/t

a = (15 i - 3j - 4 i - 1 j)/19 = 0.579 i - 0.211 j

ax = 0.579 m/s^2

ay = -0.211 m/s^2

b)theta = tan^-1 (-0.211/0.579) = -20.02

theta = 360 - 20.02 = 340 deg

Hence, theta = 340 deg

c)we know that

s = s0 + ut + 1/2 at^2

s = 16 i - 4.2 j + (4 i + 1 j) 28 + 1/2 x 28^2 (0.579 i -0.211 j)

s = 16 i - 4.2 j + 112 i + 28 j + 227 i - 82.71 j

s = 355 i - 58.91 j

x = 355 i

y = -58.91 j

theta = tan^-1(-58.91/355) = -9.4

theta = -9.4 + 360 = 350.6 deg

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