A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a p

Question

A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to a certain rock is i = (11.0 ? 2.70 ) m. After the fish swims with constant acceleration for 18.0 s, its velocity is = (18.0 ? 2.00 ) m/s.

The final location of the fish in unit-vector notion is the position vector f = 454.055555555556 i + (-43.78)j m. So it has coordinates (x,y) = (454,-43 m).

Its direction of motion is the direction of its velocity. Be sure that your final answer for ? is measured counterclockwise from the +x axis.

? = tan?1 ( vy/vx)

= tan-1 ( (m/s) / (m/s))?

=degree?

Explanation / Answer

lets calculate x coordinate and y coordinate seperately for x v=4m/s@ t=0 x=11 a=(18-4)18=0.7777 final x=454 454=4t+0.5*0.7777*t^2...(1) for y v=1 @t=0 y=2.7 a=(1-2)18=-0.0555 final y=-43 -43=1t-0.5*0.055558t^2..(2) we get t values from 1 and 2 take the value which satisfies both..=29.41sec Vx=4+29.41*0.7777=26.872 Vy=-0.6333 angle=tan-1(-0.6333/26.872)=-1.35 =1.35deg clockwise plss rate

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