Lesson 3 · 35 min

Rectangular (x–y) Components

With fixed axes, a curved motion splits into two straight-line motions, one along \(x\) and one along \(y\). Every rectilinear skill you already have now works twice over.

Learning objectives

Position, velocity and acceleration in components

Put fixed \(x\) and \(y\) axes at the origin. The position of the particle is

\[ \rvec = x\,\colX{\ihat} + y\,\colY{\jhat} \]

The unit vectors \(\ihat\) and \(\jhat\) never change size or direction, so their time derivatives are zero. Differentiating \(\rvec\) therefore only differentiates the coordinates:

Rectangular components

\[ \begin{aligned} \vvec &= \dot x\,\colX{\ihat} + \dot y\,\colY{\jhat} = v_x\,\colX{\ihat} + v_y\,\colY{\jhat} \\ \avec &= \ddot x\,\colX{\ihat} + \ddot y\,\colY{\jhat} = a_x\,\colX{\ihat} + a_y\,\colY{\jhat} \end{aligned} \]

with \(v_x = \dot x,\ v_y = \dot y,\ a_x = \dot v_x = \ddot x,\ a_y = \dot v_y = \ddot y\).

So a plane curvilinear motion is two rectilinear motions happening at the same time: the "shadow" of the particle on the \(x\)-axis moves with \(x(t)\), and its shadow on the \(y\)-axis moves with \(y(t)\). The magnitudes and directions follow from the components:

\[ v = \sqrt{v_x^2 + v_y^2}, \qquad \theta_v = \atantwo(v_y,\ v_x), \qquad |\avec| = \sqrt{a_x^2 + a_y^2} \]

Here \(\theta_v\) is the direction of travel, measured counter-clockwise from the \(+x\) axis. Because \(\vvec\) is tangent to the path, \(\tan\theta_v = v_y/v_x\) is also the slope of the path, \(dy/dx\).

Figure 3.1 A pen-plotter head draws a figure eight with \(x = 4\sin t\) and \(y = 2\sin 2t\) (metres, seconds). The dashed gray and olive arrows are \(v_x\) and \(v_y\), drawn tip to tail so they add up to \(\vvec\); the solid ones are \(a_x\) and \(a_y\). Each component is simply the derivative of one coordinate. Where the path crosses itself the head passes twice, with different velocities.

When \(x(t)\) and \(y(t)\) are given

This is the most direct case: differentiate each coordinate once for velocity and twice for acceleration, then evaluate at the time you need.

Example 3.1 — A robot gripper

The gripper of a pick-and-place robot follows \(x = 0.5t^3\ \text{m}\) and \(y = 3t^2 - 2t\ \text{m}\), with \(t\) in seconds. At \(t = 2\ \text{s}\), find its position, velocity (magnitude and direction) and acceleration.

Show solution

Position. \(x = 0.5(2)^3 = 4\ \text{m}\) and \(y = 3(2)^2 - 2(2) = 8\ \text{m}\).

Velocity. Differentiate once:

\[ \begin{aligned} v_x &= \dot x = 1.5t^2 = 6\ \text{m/s} \\ v_y &= \dot y = 6t - 2 = 10\ \text{m/s} \end{aligned} \] \[ v = \sqrt{6^2 + 10^2} = \sqrt{136} \approx 11.66\ \text{m/s}, \qquad \theta_v = \arctan\tfrac{10}{6} \approx 59.04^\circ \]

(both components are positive, so the arctangent is already in the right quadrant).

Acceleration. Differentiate again:

\[ a_x = \ddot x = 3t = 6\ \text{m/s}^2, \qquad a_y = \ddot y = 6\ \text{m/s}^2 \] \[ |\avec| = \sqrt{6^2 + 6^2} = 6\sqrt2 \approx 8.485\ \text{m/s}^2 \ \text{at}\ 45^\circ \]

Interpret. \(\avec\) (at \(45^\circ\)) is not parallel to \(\vvec\) (at \(59^\circ\)), so the gripper is both speeding up and turning, as Lesson 2 predicted.

When the path \(y = f(x)\) is given

Often you know the shape of the path, for example the profile of a guide, a cam or a road, together with one of the velocity components. Because \(x\) changes with time, \(y = f(x)\) depends on time through \(x\). Differentiate with the chain rule:

Motion along a known path \(y = f(x)\)

\[ \begin{aligned} v_y &= \dot y = f'(x)\,\dot x = f'(x)\,v_x \\ a_y &= \dot v_y = f''(x)\,v_x^2 + f'(x)\,a_x \end{aligned} \]

The second line uses the product rule on \(f'(x)\,v_x\): \(\tfrac{d}{dt}f'(x) = f''(x)\,\dot x\).

The first line says again that \(\vvec\) is tangent to the path: \(v_y/v_x = f'(x)\) is the slope. The second line holds a surprise: even when \(v_x\) is constant (\(a_x = 0\)), \(a_y = f''(x)\,v_x^2\) is not zero wherever the path is curved.

Figure 3.2 A slider moves along the guide \(y = 0.1x^2\) (metres) with a constant horizontal velocity \(v_x\). There is no \(a_x\), yet the acceleration \(a_y = f''(x)\,v_x^2 = 0.2\,v_x^2\) points up everywhere: the same value at every point of this parabola. Double \(v_x\) and \(a_y\) becomes four times as large.

Example 3.2 — A slider on a parabolic guide

The slider in Figure 3.2 moves with a constant \(v_x = 4\ \text{m/s}\) along \(y = 0.1x^2\). Find its velocity and acceleration when \(x = 5\ \text{m}\).

Show solution

Derivatives of the path. \(f'(x) = 0.2x\) and \(f''(x) = 0.2\ \text{m}^{-1}\). At \(x = 5\): \(f' = 1\).

Velocity. \(v_y = f'(x)\,v_x = (1)(4) = 4\ \text{m/s}\), so \(\vvec = 4\ihat + 4\jhat\ \text{m/s}\), with \(v = 4\sqrt2 \approx 5.657\ \text{m/s}\) at \(45^\circ\), along the tangent (slope 1).

Acceleration. \(a_x = 0\) because \(v_x\) is constant, and

\[ a_y = f''(x)\,v_x^2 + f'(x)\,a_x = (0.2)(4)^2 + (1)(0) = 3.2\ \text{m/s}^2 \]

So \(\avec = 3.2\,\jhat\ \text{m/s}^2\): straight up, toward the concave side of the guide.

Working backwards: from acceleration to position

When the acceleration is known (for example, from the forces through Newton's second law), integrate each component separately, using the initial conditions to fix the constants:

\[ v_x = v_{x0} + \int_0^t a_x\,dt, \qquad x = x_0 + \int_0^t v_x\,dt \qquad \text{(and the same for } y\text{)} \]

Example 3.3 — A delivery drone takes off

A drone starts from rest at the origin. Its acceleration is \(\avec = 1.2\,\ihat + (3 - 0.6t)\,\jhat\ \text{m/s}^2\). Find its velocity and position at \(t = 4\ \text{s}\).

Show solution

Velocity. Integrate each component from \(\vvec(0) = \mathbf{0}\):

\[ \begin{aligned} v_x &= \int_0^t 1.2\,dt = 1.2t \\ v_y &= \int_0^t (3 - 0.6t)\,dt = 3t - 0.3t^2 \end{aligned} \]

Position. Integrate again from \(\rvec(0) = \mathbf{0}\):

\[ x = 0.6t^2, \qquad y = 1.5t^2 - 0.1t^3 \]

At \(t = 4\ \text{s}\).

\[ \begin{aligned} \vvec &= 4.8\,\ihat + 7.2\,\jhat\ \text{m/s}, \quad v \approx 8.653\ \text{m/s at } 56.31^\circ \\ \rvec &= 9.6\,\ihat + 17.6\,\jhat\ \text{m} \end{aligned} \]

Check. Differentiating \(y = 1.5t^2 - 0.1t^3\) twice gives \(3 - 0.6t\), the given \(a_y\). ✓ Note that \(a_y\) is not constant, so \(y = \tfrac12 a t^2\) would have been wrong.

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Key takeaways