Question

A duck swimming on the surface of a pond has an

Answer 1

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:

$\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}$

The acceleration is:

$\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}$

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:

$v^{\to} = u ^{\to} + a^{\to} t$

Then, we can replace their values into the equation of motion in order to determine the speed:

$\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}$

$\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}$

$\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}$

Collect like terms:

$\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0 )\hat x+( 0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}$

$\mathbf{v^{\to} =0.87500 \hat x- 1.01629 \hat y}$

Thus, the magnitude is:

$\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}$

$\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}$

$\mathbf{v^{\to} =\sqrt{1.79848}}$

$\mathbf{v^{\to} =1.34 \ m/s}$

Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s

Learn more about vectors here:

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