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Question: A squirrel starts at a spot on the ground just below a bird feeder. The squirrel moves \[10\,m\] nor...

A squirrel starts at a spot on the ground just below a bird feeder. The squirrel moves 10 m10\,m north, then 7 m7\,m west, then 6 m6\,m south, then 4 m4\,m east. Where does the squirrel end up, relative to the spot where he started?

Explanation

Solution

Calculate the net horizontal and vertical movement of the squirrel by specifying the positive and negative axes.Also please remember the concept of kinematics.

Complete step by step answer:
The net displacement of the squirrel is,
S⃗=xi^+yj^\vec S = x\hat i + y\hat j

Here, xxis the total displacement along the x direction and y is the total displacement along y direction.

We need to set the positive and negative direction for the motion of the squirrel. Suppose, East is positive direction, west is negative direction, north is positive direction and south is negative direction.

Therefore, the total displacement of the squirrel is,
S⃗=(xEast−xWest)i^+(yNorth−ySouth)j^\vec S = \left( {{x_{East}} - {x_{West}}} \right)\hat i + \left( {{y_{North}} - {y_{South}}} \right)\hat j

Substitute 4 m4\,m for xEast{x_{East}}, 7 m7\,m for xWest{x_{West}}, 10 m10\,m for yNorth{y_{North}} and 6 m6\,m for ySouth{y_{South}} in the above equation.
S⃗=(4 −7)i^+(10−6)j^\vec S = \left( {4\, - 7} \right)\hat i + \left( {10 - 6} \right)\hat j
⇒S⃗=−3i^+4j^\Rightarrow \vec S = - 3\hat i + 4\hat j

The magnitude of the net displacement of the squirrel from the spot where he started is,
∣S⃗∣=x2+y2\left| {\vec S} \right| = \sqrt {{x^2} + {y^2}}

Substitute −3 - 3 for x and 4 for y in the above equation.
∣S⃗∣=(−3)2+(4)2\left| {\vec S} \right| = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( 4 \right)}^2}}
⇒∣S⃗∣=25\Rightarrow \left| {\vec S} \right| = \sqrt {25}
∴∣S⃗∣=5 m\therefore \left| {\vec S} \right| = 5\,m

Therefore, the total displacement of the squirrel from the spot where he started is 5 meters.

Now, we need to determine the position of the squirrel at the end.

The direction of the squirrel is,
θ=tan⁡−1(xy)\theta = {\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)

Substitute 3 for x and 4 for y in the above equation.
θ=tan⁡−1(34)\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)
⇒θ=tan⁡−1(0.75)\Rightarrow \theta = {\tan ^{ - 1}}\left( {0.75} \right)
∴θ=39∘\therefore \theta = 39^\circ

Therefore, the squirrel will halt at 5 m5\,m from the spot where he started towards northwest.

Note:
While solving the vector algebra, the direction of the motion is important. Generally, East is positive direction, west is negative direction, north is positive direction and south is negative direction