Respuesta :
From the given information:
- Taking the movement of the Duck in the North as the x-direction
- The movement of the Duck in the East direction as the y-direction
However, we will have to compute the initial velocity and the acceleration of the duck in their vector forms.
In vector form;
The initial velocity is:
[tex]\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}[/tex]
The acceleration is:
[tex]\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}[/tex]
The objective of this question is to determine the speed of the duck at a certain time. Since it is not given, let's assume we are to determine the Duck speed after 4 seconds of accelerating;
Then, it implies that time (t) = 4 seconds.
Using the first equation of motion:
[tex]v^{\to} = u ^{\to} + a^{\to} t[/tex]
Then, we can replace their values into the equation of motion in order to determine the speed:
[tex]\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}[/tex]
[tex]\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}[/tex]
[tex]\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}[/tex]
Collect like terms:
[tex]\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0 )\hat x+( 0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}[/tex]
[tex]\mathbf{v^{\to} =0.87500 \hat x- 1.01629 \hat y}[/tex]
Thus, the magnitude is:
[tex]\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}[/tex]
[tex]\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}[/tex]
[tex]\mathbf{v^{\to} =\sqrt{1.79848}}[/tex]
[tex]\mathbf{v^{\to} =1.34 \ m/s}[/tex]
Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s
Learn more about vectors here:
https://brainly.com/question/17108011?referrer=searchResults