A fish swimming in a horizontal plane has velocity v with arrowi = (4.00 i + 1.0

Question

A fish swimming in a horizontal plane has velocity v with arrowi = (4.00 i + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is r with arrowi = (16.0 i 1.80 j) m. After the fish swims with constant acceleration for 19.0 s, its velocity is v with arrow = (19.0 i 6.00 j) m/s.

A) what are the components of the acceleratioin of the fish?

ax = ____ m/s2

ay = ____ m/s2

_______ ° counterclockwise from the +x-axis

C) If the fish maintains constant acceleration, where is it at t = 26.0 s?

x = ______ m

y= _______ m

In what direction is it moving?

______  ° counterclockwise from the +x-axis

Explanation / Answer

here,

initial velocity , u = ( 4 i + 1 j) m/s

initial position , x0 = ( 16 i - 1.8 j) m/s

final velocity , v = ( 19 i - 6 j) m/s

A)

time taken , t = 19 s

accelration , a = (v - u)/t

a = (15 i - 7 j )/19 m/s^2

a = ( 0.79 i - 0.37 j) m/s^2

ax = 0.79 i m/s^2 , ay = - 0.37 j m/s^2

B)

theta = arctan( 0.37/0.79)

theta = 25.1 degree

theta = ( 180 - 25.1)

the angle is 154.1 degree counterclockwise from +x axis

C)

after 26 s,

x - x0 = u * t + 0.5 * a * t^2

( x - 16 i +1.8 j) = ( 4i + 1 j) * 26 + 0.5 * ( 0.79 i - 0.37 j ) *26^2

x = (387.02 i - 124.26 j) m

x = 387.02 m

y = 125.86 m

D)

theta = arctan( y/x) = arctan( 125.86/387.02)

theta = 18.01 degree

as theta is in fourth quadrant

the direction is 341.99 degree counterclockwise from the +x axis

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