A fish swimming in a horizontal plane has velocity v_i = (4.00 i - 3.00 j) m/s a

Question

A fish swimming in a horizontal plane has velocity v_i = (4.00 i - 3.00 j) m/s at a point in the ocean where the position relative to a certain rock is r_i = (-10.0 i - 4. After the fish swims with constant acceleration for 24.0 s, its velocity is v = (16.0 i - 5.00 j) m/s. What are the components of the acceleration? What is the direction of the acceleration with respect to unit vector i? (counterclockwise from the +x-axis is positive) If the fish maintains constant acceleration, where is it at t = 32.0 s, with respect to the rock? In what direction is it moving? (counterclockwise from the +x-axis is positive)

Explanation / Answer

a) ax = (v2x - v1x)/t

= (16 - 4)/24

= 0.5 m/s^2

ay = (v2y - v1y)/t

= (-5 - (-3))/24

= -0.083 m/s^2

direction : theta = tan^-1(ay/ax)

= tan^-1(-0.083/0.5)

= 9.46 degrees below +x axis

= 350.54 degrees counterclockwise from +x axis

b)

x = xo + v1x*t + 0.5*ax*t^2

= -10 + 4*32 + 0.5*0.5*32^2

= 374 m

y = yo + v1y*t + 0.5*ay*t^2

= -4 + (-3)*32 + 0.5*(-0.083)*32^2

= 142.5 m

c) vfx = v1x + ax*t

= 4 + 0.5*32

= 20 m/s

vfy = v1y + ay*t

= -3 + (-0.083)*32

= -5.66 m/s

theta = tan^-1(vfy/vfx)

= tan^-1(-5.66/20)

= 15.6 degree below +x axis

= 344.4 degrees counterclockwise from +x axis

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