Physics, asked by Leem9856, 9 months ago

Leslie walks 20m north, 35m east, 10m north, and then 6m west before stopping. What is her displacement from her original location?

Answers

Answered by Anonymous
18

Explanation:

Displacement is a vector quantity; it will have both magnitude and direction.

First we need to find his total distance travelled along the y-axis. Let's say that all of her movement north is positive and south is negative.

∑Y=20m+10m=30m. She moved a net of 30 meters to the north.

Now let's do the same for the x-axis, using positive for east and negative for west.

∑X=35m−6m=29m. She moved a net of 29 meters to the east.

Now, to find the resultant displacement, we use the Pythagorean Theorem. Her net movement north will be perpendicular to her net movement east, forming a right triangle. Her location relative to her starting point will be the hypotenuse of the triangle.

D2=(∑X)2+(∑Y)2

D2=(29m)2+(30m)2

D2=841m2+900m2

D2=1741m2

Now take the square root of both sides.

√D2=√1741m

D=41.73m

Since we are solving for a vector, we also need to find the direction of this distance. We do this by solving for the angle of displacement.

To find the angle, we use the arctan of our directional displacements in the x- and y-axes. The tangent of the angle will be equal to the x-displacement over the y-displacement.

tan(Θ)=∑X∑Y

Θ=tan−1(∑X∑Y)

\Theta =\tan^{-1}({\frac{29m}{30m})

Θ=44.03∘

Combining the magnitude and direction of our distance gives us the displacement: 41.73m at 44.3∘.

Answered by Anonymous
2

Explanation:

nakzmanzhzkaznjzkaznz

Similar questions