Question

A duck has a mass of 2.5 kg. As the duck paddles, a force of 0.15 N acts on it in a direction due east.
In addition, the current of the water exerts a force of 0.25 N in a direction 50° south of east. When
these forces begin to act, the velocity of the duck is 0.12 m/s in a direction due east. Find the
magnitude and direction of the displacement that the duck undergoes in 4 sec. while the forces are
acting.

Answer 1

Force East = 0.1 + (0.2 . cos 52º) = 0.223 N 

Force South = 0.2 . sin 52º = 0.158 N 

Total force = √ (0.223^2 + 0.158^2) = 0.273 N 

Acceleration = 0.273 / 2.5 = 0.109 m/s/s 

both at direction E 35.3ºS

Answer 2

Concept Introduction:-

It could take the shape of a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that a duck has a mass of $2.5$ kg. As the duck paddles, a force of $0.15$ N acts on it in a direction due east. In addition, the current of the water exerts a force of $0.25$ N in a direction $50\°$ south of east. When these forces begin to act, the velocity of the duck is $0.12$ m/s in a direction due east.

To Find:-

We have to find that the magnitude and direction of the displacement that the duck undergoes in $4$ sec. while the forces are acting.

Solution:-

According to the problem

Force East$= 0.1 + (0.2 \times cos 52\º) = 0.223 N$

Force South$= 0.2 \times sin 52\º = 0.158 N$

Total force$=\sqrt{(0.223^2 + 0.158^2)}= 0.273 N$

Acceleration $= 0.273 / 2.5 = 0.109$ m/s/s

both at direction E $35.3\ºS$

Final Answer:-

The correct answer is both at direction E $35.3\ºS$.

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