The component of final velocity are 7.87 m/s and -1.94 m/s respectively.
The final speed of the speedboat is 8.10 m/s.
Given data:
The initial velocity vector along x-direction is, [tex]v_{1x}=8.57 \;\rm m/s[/tex].
The initial velocity vector along y-direction is, [tex]v_{1y}=-2.61 \;\rm m/s[/tex].
The time interval is, t = 6.67 s.
The average acceleration along x-direction is, [tex]a_{x}=-0.105 \;\rm m/s^{2}[/tex].
The average acceleration along y-direction is, [tex]a_{y}=0.101 \;\rm m/s^{2}[/tex].
The concept of acceleration is used to solve the given problem. Acceleration is defined as the rate of change of velocity. So, along the x-direction it is given as,
[tex]a_{x}=\dfrac{v_{2x}-v_{1x}}{t} \\\\-0.105=\dfrac{v_{2x}-8.57}{6.67}\\\\v_{2x}= (-0.105 \times 6.67) + 8.57\\\\v_{2x}=7.87 \;\rm m/s[/tex]
Similarly, the acceleration along the y-direction is,
[tex]a_{y}=\dfrac{v_{2y}-v_{1y}}{t} \\\\0.101=\dfrac{v_{2y}-(-2.61)}{6.67}\\\\v_{2y}= (0.101 \times 6.67) - 2.61\\\\v_{2y}=-1.94 \;\rm m/s[/tex]
Thus, we can conclude that the component of final velocity are 7.87 m/s and -1.94 m/s respectively.
Now, the speedboat's final speed is calculated as,
[tex]v_{final}=\sqrt{v^{2}_{2x}+v^{2}_{2y}}\\\\v_{final}=\sqrt{7.87^{2}+(-1.94)^{2}}\\\\v_{final}=8.10 \;\rm m/s[/tex]
Thus, the final speed of the speedboat is 8.10 m/s.
Learn more about the acceleration here:
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