Lesson 1 · 20 min

From Straight Lines to Curves

A car on a highway on-ramp, a ball thrown to a teammate and a package dropped from a drone all move along curved paths. This lesson reviews straight-line motion, shows what is new when the path bends, and introduces the two coordinate systems you will use for the rest of the module.

Learning objectives

Motion along a straight line: a quick review

In rectilinear motion a particle moves along a straight line, and one number, its position \(s\) measured from a fixed origin on the line, tells you where it is. The velocity is the rate of change of position, and the acceleration is the rate of change of velocity:

Rectilinear kinematics

\[ v = \frac{ds}{dt} = \dot s, \qquad a = \frac{dv}{dt} = \dot v = \ddot s, \qquad a\,ds = v\,dv \]

A dot means a derivative with respect to time. The third relation comes from eliminating \(dt\) between the first two; use it when \(a\) is given as a function of position.

When \(a = a_c\) is constant, integrating these relations once and twice gives the familiar results (with \(s_0\) and \(v_0\) the values at \(t = 0\)):

\[ \begin{aligned} v &= v_0 + a_c t \\ s &= s_0 + v_0 t + \tfrac12 a_c t^2 \\ v^2 &= v_0^2 + 2a_c(s - s_0) \end{aligned} \]

Example 1.1 — Braking to a stop

A car travelling at \(25\ \text{m/s}\) (90 km/h) brakes with a constant deceleration of \(5\ \text{m/s}^2\). How far does it travel before it stops, and how long does that take?

Show solution

Set up. Take \(s\) positive in the direction of travel, so \(v_0 = 25\ \text{m/s}\) and \(a_c = -5\ \text{m/s}^2\) (the car slows down). It stops when \(v = 0\).

Distance. Time is not given or asked for first, so use \(v^2 = v_0^2 + 2a_c(s - s_0)\):

\[ 0 = 25^2 + 2(-5)(s - 0) \quad\Rightarrow\quad s = \frac{625}{10} = 62.5\ \text{m} \]

Time. From \(v = v_0 + a_c t\): \(0 = 25 - 5t\), so \(t = 5\ \text{s}\).

Check. The average speed while braking at a constant rate is \((25 + 0)/2 = 12.5\ \text{m/s}\), and \(12.5 \times 5 = 62.5\ \text{m}\). ✓

When the path curves

In curvilinear motion the particle moves along a curved path. When the whole path lies in one plane, as it does for a car on flat ground, a thrown ball or a point on a spinning wheel, the motion is plane curvilinear motion. That is the subject of this module; 3D paths use the same ideas with one more coordinate.

Two things change when the path curves:

Try it below. The car drives round a hairpin at a constant speed: a straight, a half circle, then another straight. Watch the red acceleration arrow.

Figure 1.1 A car rounds a hairpin (radius \(3\ \text{m}\), drawn small so it fits) at constant speed. Press Play, or drag the car. On the straights the acceleration is zero. On the curve the speed is unchanged, yet the red arrow points toward the center of the curve with size \(v^2/R\). Raise the speed: doubling \(v\) makes the acceleration four times as large.

Notice two things. First, the green velocity arrow is always tangent to the path: it points the way the car is going. Second, on the curve the acceleration is perpendicular to the velocity, pointing into the bend. Lessons 5–6 explain both facts, and show that this acceleration is exactly \(v^2/R\).

Two ways to describe the same motion

A vector can be split into components along any two perpendicular directions. In plane curvilinear motion two choices are especially useful:

Rectangular coordinates \((x, y)\)

Fixed axes, like a surveyor's grid or the view from a camera mounted on the ground. The unit vectors \(\ihat\) and \(\jhat\) never change direction.

\(\vvec = v_x\ihat + v_y\jhat\), \(\avec = a_x\ihat + a_y\jhat\). Lessons 3–4.

Path coordinates \((t, n)\)

Directions attached to the moving particle, like the driver's view: \(\et\) points forward along the path and \(\en\) points sideways, toward the inside of the curve.

\(\vvec = v\,\et\), \(\avec = a_t\et + a_n\en\). Lessons 5–7.

Neither system is "more correct": they describe the same vectors. Choose the one that makes the problem easy.

Figure 1.2 A ball thrown from a height of \(1\ \text{m}\). Switch Components between rectangular and path. The red acceleration \(\avec\) is the same arrow either way (gravity, \(9.81\ \text{m/s}^2\) straight down); only the way it is split changes. In \(x\)–\(y\) it is all \(a_y\). In \(n\)–\(t\) part of it slows the ball along the path (\(a_t\)) and part of it bends the path (\(a_n\)).

Check your understanding

Key takeaways