Question

Your dog is running around the grass in your backyard. He undergoes successive displacements, 3.60 m south, 8.46 m northeast, and 15.6 m west. What is the resultant displacement? (take + x direction to be east). Find magnitud? Find direction?

Answer 1

First, we resolve the northeast displacement into its north and east components. The angle from the positive x-axis of a northeast displacement is 45 degrees. Thus:
North = 8.46sin(45) = 5.98 m
East = 8.46cos(45) = 5.98 m

North displacement = 5.98 - 3.6 = 2.38 m
West displacement = 15.6 - 5.98 = 9.62

Magnitude = √(2.38² + 9.62²)
Magnitude = 9.91 m

Direction:
tan∅ = 2.38 / 9.62
∅ = 13.9° north from east

Answer 2

The magnitude of resultant displacement of the dog is $\boxed{9.91\,{\text{m}}}$  and the direction of resultant displacement is $\boxed{13.89^\circ }$  north of west.

Further Explanation:

The displacement of the dog is as shown in the figure attached below. The dog moves from the point $A\to B\to C\to D$ .

As the dog moves from A to B, it is a straight line and it is displaced by a distance of $3.6\,{\text{m}}$  from its initial position. After that, the dog moves by a distance of $8.46\,{\text{m}}$  in the north-east direction.

The resultant displacement of the dog from point B to point C in the horizontal direction is:

$\begin{aligned}B{C_x}&=8.46\cos45^\circ\\&=5.98\,{\text{m}}\\\end{aligned}$

The displacement from B to C in vertical direction is:

$\begin{aligned}B{C_y}&=8.46\sin45^\circ\\&=5.98\,{\text{m}}\\\end{aligned}$

Then after this displacement, the dog moves from point C to point D in the horizontal direction.

So the final displacement of the dog in the horizontal direction is:

$\begin{aligned}{d_x}&=15.6-5.98\\&=9.62\,{\text{m}}\\\end{aligned}$

Displacement of the dog in the vertical direction is:

$\begin{aligned}{d_y}&=5.98-3.6\\&=2.38\,{\text{m}}\\\end{aligned}$

Thus, the magnitude of total displacement of the dog from initial position is:

$\begin{aligned}AD&=\sqrt{{{\left({9.62}\right)}^2}+{{\left({2.38}\right)}^2}}\\&=\sqrt{98.21}\\&=9.91\,{\text{m}}\\\end{aligned}$

The direction of final displacement of the dog is:

$\begin{aligned}\theta&={\tan ^{-1}}\left({\frac{{{d_y}}}{{{d_x}}}}\right)\\&={\tan^{-1}}\left({\frac{{2.38}}{{9.62}}}\right)\\&=13.89^\circ\,{\text{north}}\,{\text{of}}\,{\text{west}}\\\end{aligned}$

Thus, the magnitude of resultant displacement of the dog is $\boxed{9.91\,{\text{m}}}$  and the direction of resultant displacement is $\boxed{13.89^\circ }$  north of west.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Distance and Displacement

Keywords:

Dog is running, grass, in your backyard, successive displacements, 3.60m south, 8.46m north east, 15.6 m west, resultant displacement, horizontal direction, vertical direction.