Vectors in Two Dimensions

< Return

Now that we are moving in two-dimensions, we need to deal with how we treat direction. We know that most quantities need direction, like velocity and acceleration.

Two-dimensional motion.

For this one-dimensional case, a simple flip of the signs works. For free falling objects, we almost always set "going up" as positive velocity and "going down" as negative velocity.

These quantities that need direction are called vectors, while those that don't need direction are called scalars. For example, velocity is a vector, while speed is a scalar.

The velocity of a person running around in a circle changes, but their speed is constant.

The figure above shows a person running around in a circle. We can see that their velocity is changing direction, and thus changes. However, the magnitude doesn't change, thus their speed is constant.

Question 1

(HRK Q2.6) Can the speed of a particle ever be negative? If so, give an example; if not, explain why.

Answer

No, it can't. Velocity can be negative, as we can either go left and right or go up and down, however speed must always be positive.

In two dimensions, we can't rely on changing signs. For example, in one-dimension, we can add vectors by just adding their magnitudes up. The vector we get from adding these up is called the resultant.

Graphically adding one-dimensional vectors.

To add two-dimensional vectors however, we have to do a different way. Visually, we just add them up tip-to-tail.

Graphically adding two-dimensional vectors.

To add them numerically, we first need to break down the vectors. A vector can be broken down into its components. To get these components, we need to use trigonometry with the angle of the vector.

A vector can be broken down into its components.

One must recall their trigonometry to deal with this; especially the sin() and cos() functions.

We can also use these components to get the magnitude and the angle of the vector via the pythagorean theorem and using the arctan() function respectively.

Exercise 1

(HRK E2.4a) What are the components of a vector

\vec{a}

in the xy plane if its direction is 252° counterclockwise from the positive x axis and its magnitude is 7.34 units?

Solution

The diagram below shows the vector. We then calculate its components.

The vector in the xy plane.

The reference angle of 252° is 72°, we take the sine and cosine of that angle to get the magnitudes of the two components.

\vec{a_{x}} = −7.34 \cos (72°)

\vec{a_{y}} = −7.34 \sin (72°)

This gives us the two component vectors.

Exercise 2

(HRK E2.3a) Vector

\vec{a}

has a magnitude of 5.2 units and is directed east. Vector

\vec{b}

has a magnitude of 4.3 units and is directed 35° west of north. Find

\vec{a} + \vec{b}

Solution

The two vectors added up in the xy plane.

We take the sine and cosine of the angle to get the two components of the angled vector.

\vec{b_{x}} = −4.3 \sin (35°) = −2.47

\vec{b_{y}} = 4.3 \cos (35°) = 3.52

To get the magnitude, we use the pythagorean theorem.

R = \sqrt{[3.52]² + [5.2 - 2.47]²} = 4.45

For the angle measure, we use the arctan() function.

θ = arctan(\frac{3.52}{5.2 - 2.47}) = 52.20°

Thus, we get that $\vec{R}$ is a vector with magnitude 4.45 units and direction 52.20° counterclockwise from the +x axis.

One way to represent a vector is by using unit vectors. Here, we multiply a unit vector by the magnitude of the components. This effectively gives us the length and height of any given vector.

A vector, defined by its unit vectors.

Exercise 3

(LSM P4.18) The position of a particle changes from

\vec{r_{1}} = (2.0î + 3.0ĵ)

\vec{r_{2}} = (−4.0î + 3.0ĵ)

. What is the particle’s displacement?

Solution

We first parse what the two vectors mean. The first vector $\vec{r_{1}}$ has a length of 2 going to the right and a height of 3 going up, while the other vector r2→ has a length of 4 going to the left (hence the negative sign) and a height of 3 going up also.

The two positions.

We got these values by looking at the unit vectors they are composed out of.
To do this formally, we use the distance formula, which is powered by the pythagorean theorem.

d = \sqrt{[2.0 − (−4.0)]² + [3.0 − 3.0]²} = 6.0

One may notice that this shows that the x-component and y-component of a vector is separate; and thus we can separate motion in two dimensions to two cases of one-dimensional motion, which we've studied.

Two-dimensional motion is two cases of one-dimensional motion.

We can now deal with some problems about motion in two dimensions, where we just have to break down the vectors into its components and work at it from there.

Be warned; this may seem like a large jump in complexity as there looks to be a lot of information, however just remember one thing: two-dimensional motion is two cases of one-dimensional motion.

Problem 1

(LSM P4.18) A boat heads out into a lake with an acceleration of

(2.0î)

m/s². A strong wind is pushing the boat, giving it an additional velocity of

(2.0î + 1.0ĵ)

m/s.
(a) What is the velocity of the boat at t = 10s?
(b) What is the position of the boat at t = 10s?

Solution

The acceleration appears to only move in the x-axis, so the velocity's y-component will not be affected. We can calculate the velocity by just using our kinematic equations. To make it less confusing, let's add a subscript of $x$ and $y$ to determine what dimension we are using.

v_{x} = v_{0x} + a_{x} t

v_{x} = (2.0m/s) + (2m/s²) (10s) = 22m/s

This gives us that the velocity at 10s is $(22.0î + 1.0ĵ)$ m/s. For position, we have to separate it first into the x-component and y-component. For the x-component, we have an initial velocity, acceleration, and time, so we use our third equation.

x = x_{0} + v_{0x} t + \frac{1}{2} a_{x} t^{2}

x = (0m) + (2m/s) (10s) + \frac{1}{2} (2m/s²) {(10s)}^{2}

x = 60m

The y-component is much easier, as there is no acceleration to worry about. We use the first equation instead, as the velocity is constant.

y = y_{0} + v_{0y} t

y = (0m) + (1m/s) (10s)

y = 10m

This gives us that the position at 10s is at $(60î + 10ĵ)$ m.

Problem 2

(LSM P4.30) The acceleration of a particle is a constant. At t = 0s, the velocity of the particle is

(10î + 20ĵ)

m/s. At t = 4s, the velocity is

(10ĵ)

m/s. What is the particle’s acceleration?

Solution

The x-component of velocity has gone from 10m/s to 0m/s in four seconds. We can plug this into our second kinematic equation.

v_{x} = v_{0x} + a_{x} t

(0m/s) = (10m/s) + a_{x} (4s)

(−2.5m/s²) = a_{x}

Similarly, the y-component of velocity has gone from 20m/s to 10m/s in four seconds. Plugging this in gives us the y-component of the acceleration.

v_{y} = v_{0y} + a_{y} t

(10m/s) = (20m/s) + a_{y} (4s)

(−2.5m/s²) = a_{y}

We thus get that the acceleration vector is given by $(−2.5î + −2.5ĵ)$ m/s².

Problem 3

(LSM P4.27a) A particle’s acceleration is

(4.0î + 3.0ĵ)

m/s². At t = 0, its position and velocity are zero. What are the particle’s position and velocity as functions of time?

Solution

We must use calculus to get the velocity and position functions. Since the initial velocity and position is zero, we may drop the arbitrary constant.

Starting with the x-component, we first get the velocity then position function.

\int (4.0) dt = 4.0 t

\int (4.0 t) dt = 2.0 t^{2}

We follow a similar way for the vertical component.

\int (3.0) dt = 3.0 t

\int (3.0 t) dt = \frac{3.0 t^{2}}{2}

This gives us the velocity function is $(4.0tî + 3.0tĵ)$ m/s, while the position function is $(2.0 t^{2} î + \frac{3.0 t^{2}}{2} ĵ)$ m/s.

Vectors in Two Dimensions

Question 1

Answerv

Exercise 1

Solutionv

Exercise 2

Solutionv

Exercise 3

Solutionv

Problem 1

Solutionv

Problem 2

Solutionv

Problem 3

Solutionv

Answer

Solution

Solution

Solution

Solution

Solution

Solution