# The x- and y- component of an acceleration vector are 3.0 m/s^2 and 4.0 m/s^2, respectively. (a)...

## Question:

The x- and y- component of an acceleration vector are 3.0 m/s{eq}^2 {/eq} and 4.0 m/s{eq}^2 {/eq}, respectively.

(a) The magnitude of the acceleration vector is :

(1) less than 3.0 m.s{eq}^2 {/eq}, (2) between 3.0 m/s{eq}^2 {/eq} and 4.0 m/s{eq}^2 {/eq}, (3) between 4.0 m/s{eq}^2 {/eq} and 7.0 m/s{eq}^2 {/eq}, (4) equal to 7.0 m/s{eq}^2 {/eq}.

(b) What are the magnitude and direction of the acceleration?

## Vectors

A vector when decomposed into Cartesian coordinates can be written as $$\mathbf{V} = V_{x} \, \mathbf{u}_{x} + V_{y} \, \mathbf{u}_{y} \, \, ,$$

where {eq}\mathbf{u}_{x} {/eq} and {eq}\mathbf{u}_{y} {/eq} are the unit vectors in the {eq}x {/eq} and {eq}y {/eq} directions, respectively. In terms of the angle between the vector itself and the positive direction of {eq}x {/eq}-axis, we have

\begin{align} V_{x} &= V \cos \theta \\ \\ V_{y} &= V \sin \theta \, \, . \end{align}

The magnitude {eq}V {/eq} is given by

$$V = \sqrt{ V_{x}^{2} + V_{y}^{2} } \, \, ,$$

and the angle {eq}\theta {/eq} is given by

$$\theta = \arctan \left( \dfrac{ V_{y} }{ V_{x} } \right) \, \, .$$

Triangle Inequality The triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the remaining side.

For right triangles specifically, the hypotenuse is greater than either of the other two sides and less than their sum.

(a) The magnitude of the vector is equal to the hypotenuse of a right triangle of sides {eq}3.0 \, \text{m} / \text{s}^{2} {/eq} and {eq}4.0 \, \text{m} / \text{s}^{2} {/eq} shown in the figure.

By triangle inequality, the hypotenuse must be greater than any of the sides, and less than their sum, so the answer is:

(3) between 4.0 m/s{eq}^2 {/eq} and 7.0 m/s{eq}^2 {/eq}

(b) The vector can be written as

$$\mathbf{a} = ( 3.0 \, \text{m} / \text{s}^{2} ) \, \mathbf{u}_{x} + ( 4.0 \, \text{m} / \text{s}^{2} ) \, \mathbf{u}_{y}$$

The magnitude of the acceleration vector is

$$a = \sqrt{ ( 3.0 \, \text{m} / \text{s}^{2} )^{2} + ( 4.0 \, \text{m} / \text{s}^{2} )^{2} } = 5.0 \, \text{m} / \text{s}^{2} \, \, .$$

The angle between the vector and the positive direction of {eq}x {/eq}-axis is given by

$$\theta = \arctan \left( \dfrac{ 4.0 }{ 3.0 } \right) \approx 53^{\circ} \, \, .$$