Question

Please help pleease

Answer 1

1) The distance of the store from the house is 4.47 blocks

2) The horizontal component of the vector is 37.7 N, the vertical component is 16.0 N

3) The magnitude of the resultant vector is 11.6 m

Explanation:

1)

Jeremy's motion is divided into two parts:

- First, a motion of 4 blocks east

- Then, a motion of 2 blocks south

Here we want to measure the distance between the initial and final position of Jeremy.

We notice that the two displacements are perpendicular to each other, so they correspond to the sides of a right triangle, of which the hypothenuse is the distance between the initial and final point. Therefore, the distance can be found by using Pythagorean's theorem:

$R=\sqrt{d_1^2+d_2^2}=\sqrt{4^2+2^2}=4.47$

So, 4.47 blocks.

2)

The horizontal and vertical components of the vector can be calculated as

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v = 41 N is the magnitude of the vector

$\theta=23^{\circ}$ is the angle between the direction of the vector and the x-axis (which corresponds to the east direction)

Substituting numbers into the equations, we find

$v_x = (41)(cos 23)=37.7 N$

$v_y = (41)(sin 23)=16.0 N$

3)

In the previous part, we said that the horizontal component of a vector is given by

$v_x = v cos \theta$

where

v is the magnitude of the vector

$\theta$ is the angle

In this problem we know that:

$v_x = 11 m$ is the horizontal component

$\theta=19^{\circ}$ is the angle

Therefore, we can re-arrange the equation to find v, the magnitude of the resultant vector:

$v=\frac{v_x}{cos \theta}=\frac{11}{cos 19}=11.6 m$

Learn more about vectors here:

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