Physics, 29.10.2019 06:31 jewelz5887

Ahiker walks 200m west and then walks 100m north. what is the magnitude and direction of her resulting displacement?

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A hiker walks 200 m west and then walks 100 m north, the resulting magnitude is 223 m. The direction can be solve using trigo function sin angle = opposite/ hypotenuse. Which is equal to 26.6 degree. So the displacement and direction is 223 m 26.6 degree North of west.

Answer from: michael4443

The magnitude of displacement of hiker is $\fbox{\begin\\224\,{\text{m}}\end{minispace}}$ and the direction of resulting displacement is $\fbox{north of west direction}$ .

Further explanation:

The displacement is the shortest distance travelled by the object from its initial position to the final destination.

Given:

The distance of hiker in the west direction is $200\,{\text{m}}$ .

The distance of hiker in the north direction is $100\,{\text{m}}$ .

Concept used:

The hiker start from point A, moves to the west direction and reaches point B then she moves 100m in the north direction and reaches point C. Now, we need to find the shortest distance that is the displacement of the hiker. The distance from point C to point A will be the shortest distance of the hiker.

The hiker moves along a right angled triangle ABC in which AB is treated as the base of the triangle, BC is treated as the altitude of the triangle and CA is treated as the hypotenuse of the triangle.

According to Pythagoras theorem “the square of the longest side is the sum of the base and altitude of the right angled triangle.”

The expression for the Pythagoras theorem is given as.

${\left( {AC} \right)^2} = {\left( {BC} \right)^2} + {\left( {AB} \right)^2}$

Rearrange the above expression for the hypotenuse.

$\left( {AC} \right) = \sqrt {{{\left( {BC} \right)}^2} + {{\left( {AB} \right)}^2}}$ …… (1)

In the angle between side AB and AC is given as.

$\theta={\sin ^{-1}}\left( {\dfrac{{BC}}{{AC}}} \right)$ …… (2)

Substitute $200\,{\text{m}}$ for AB and $100\,{\text{m}}$ for BC in equation (1).

$\begin{aligned}\left( {AC}\right)&=\sqrt {{{\left( {200\,{\text{m}}} \right)}^2} + {{\left( {100\,{\text{m}}} \right)}^2}}\\&=223.6\,{\text{m}}\\&\simeq {\text{224}}\,{\text{m}} \\ \end{aligned}$

Substitute $100\,{\text{m}}$ for BC and $224\,{\text{m}}$ for AC in equation (2).

$\begin{aligned}\theta&={\sin ^{-1}}\left( {\frac{{100\,{\text{m}}}}{{224\,{\text{m}}}}}\right) \\&={\left( {26.51} \right)^ \circ }\\ \end{aligned}$

Thus, the shortest distance of the hiker is $224\,{\text{m}}$ in the $\fbox{North West direction}$ at an angle of ${26.51^ \circ }$ .

Learn more:

1. Conservation of momentum .

2. Motion under friction .

3. Motion under gravitation .

Answer Details:

Grade: High School

Subject: Physics

Chapter: Speed and distance

Keywords:

Speed, distance, displacement, velocity, time, magnitude, direction, north, east, west, south, Pythagoras theorem, triangle, sine, cosine, tangent, 224 m, 223.6 m, 26.51 deg, 27 deg.

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