4.1 Motion in Two Dimensions

Motion diagrams are an important tool for visualizing motion, and we’ll continue to use them, but we also need to develop a mathematical description of motion in two dimensions. For convenience, we’ll say that any two-dimensional motion is in the xy-plane regardless of whether the plane of motion is horizontal or vertical.

FIGURE 4.1 shows a particle moving along a curved path—its trajectory—in the xy-plane. We can locate the particle in terms of its position vector $\vec{r} = x î + y ĵ$ .

The figure illustrates a particle moving along a curved path in the x y coordinate plane.

FIGURE 4.1 Full Alternative Text

NOTE In Chapter 2 we made extensive use of position-versus-time graphs, either x versus t or y versus t. Figure 4.1, like many of the graphs we’ll use in this chapter, is a graph of y versus x. In other words, it’s an actual picture of the trajectory, not an abstract representation of the motion.

FIGURE 4.2a shows the particle moving from position $\vec{r}_{1}$ at time $t_{1}$ to position $\vec{r}_{2}$ at a later time $t_{2}$ . The average velocity—pointing in the direction of the displacement $Δ \vec{r}$ —is

\vec{v}_{avg} = \frac{Δ \vec{r}}{Δ t} = \frac{Δ x}{Δ t} î + \frac{Δ y}{Δ t} ĵ

(4.1)

The figure consists of three parts labeled ("a"), (b), and (c) that illustrate the average velocity and the instantaneous velocity of an object at a point in projectile motion using the x y coordinate plane.

FIGURE 4.2 Full Alternative Text

You learned in Chapter 2 that the instantaneous velocity is the limit of $\vec{v}_{avg}$ as $Δ t \to 0$ . As $Δ t$ decreases, point 2 moves closer to point 1 until, as FIGURE 4.2b shows, the displacement vector becomes tangent to the curve. Consequently, the instantaneous velocity vector $\vec{v}$ is tangent to the trajectory.

Mathematically, the limit of Equation 4.1 gives

\vec{v} = lim_{Δ t \to 0} \frac{Δ \vec{r}}{Δ t} = \frac{d \vec{r}}{d t} = \frac{d x}{d t} î + \frac{d y}{d t} ĵ

(4.2)

We can also write the velocity vector in terms of its x- and y-components as

\vec{v} = v_{x} î + v_{y} ĵ

(4.3)

Comparing Equations 4.2 and 4.3, you can see that the velocity vector $\vec{v}$ has x- and y-components

v_{x} = \frac{d x}{d t} and v_{y} = \frac{d y}{d t}

(4.4)

That is, the x-component $v_{x}$ of the velocity vector is the rate $d x / d t$ at which the particle’s x-coordinate is changing. The y-component is similar.

FIGURE 4.2c illustrates another important feature of the velocity vector. If the vector’s angle $θ$ is measured from the positive x-direction, the velocity vector components are

\begin{array}{l} v_{x} = v cos θ \\ v_{y} = v sin θ \end{array}

(4.5)

where

v = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}

(4.6)

is the particle’s speed at that point. Speed is always a positive number (or zero), whereas the components are signed quantities (i.e., they can be positive or negative) to convey information about the direction of the velocity vector. Conversely, we can use the two velocity components to determine the direction of motion:

θ = {tan}^{- 1} (\frac{v_{y}}{v_{x}})

(4.7)

NOTE In Chapter 2, you learned that the value of the velocity is the slope of the position-versus-time graph. Now we see that the direction of the velocity vector $\vec{v}$ is the tangent to the y-versus-x graph of the trajectory. FIGURE 4.3, reminds you that these two graphs use different interpretations of the tangent lines. The tangent to the trajectory does not tell us anything about how fast the particle is moving.

The figure illustrates two parts that depict the different uses of tangent lines for graphs of one-dimensional and two-dimensional motion.

FIGURE 4.3 Full Alternative Text

EXAMPLE 4.1 Finding velocity

A sports car’s position on a winding road is given by

\vec{r} = (6.0 t - 0.10 t^{2}) î + (8.0 t - 0.00095 t^{3}) ĵ

where the y-axis points north, t is in s, and r is in m. What are the car’s speed and direction at $t = 120 s ?$

MODEL Model the car as a particle.

SOLVE Velocity is the derivative of position, so

\begin{array}{l} v_{x} & = & \frac{d x}{d t} = 6.0 - 2 (0.10 t) \\ v_{y} & = & \frac{d y}{d t} = 8.0 - 3 (0.00095 t^{2}) \end{array}

Written as a vector, the velocity is

\vec{v} = (6.0 - 0.20 t) î + (8.0 - 0.00285 t^{2}) ĵ

where t is in s and v is in m/s. At $t = 120 s$ , we can calculate $\vec{v} = (- 18 î - 33 \hat{j}) m/s$ . The car’s speed at this instant is

v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{{(- 18 m/s)}^{2} + {(- 33 m/s)}^{2}} = 38 m/s

Both components of the velocity vector are negative, so the direction of motion is to the left (west) and down (south). The angle below the negative x-axis is

θ = {tan}^{- 1} (\frac{| - 33 m/s |}{| - 18 m/s |}) = 61 °

So, at this instant, the car is headed 61° south of west at a speed of 38 m/s.

4.1 Motion in Two Dimensions

FIGURE 4.1 A particle moving along a trajectory in the xy-plane.

FIGURE 4.2 The instantaneous velocity vector is tangent to the trajectory.

FIGURE 4.3 Two different uses of tangent lines.

EXAMPLE 4.1 Finding velocity