The Timpson Lyceum · University Physics I (2025)

Motion Along a Straight Line – Lecture 2 (University Physics I)

Overview

This lecture develops the language of motion in one dimension — along a straight line. It defines position, displacement, velocity, and acceleration precisely (as derivatives and integrals of one another), connects them to the shape of a position-vs-time graph, and then derives the four kinematic equations that govern any situation with constant acceleration. The chapter closes with free fall, the most important constant-acceleration case. Mastering equation selection here — choosing the right kinematic equation from what you already know — is the core skill for the rest of mechanics.

Key Concepts

Position and time are relative. There is no absolute position or time in Newtonian mechanics; you describe an event only relative to a chosen reference frame — a choice of where position zero is, when time zero is, and what units (meters, seconds) you use. The physics is the same in any frame, so pick the most convenient one (usually: you are at $x = 0$ , and $t = 0$ is when the event of interest begins).

Displacement vs. distance.

Displacement $\Delta x = x_2 - x_1$ (final minus initial) — has a sign/direction, can be negative.
Distance is the total path length traveled — always positive. Walk 2 m out and 2 m back: distance $= 4$ m, displacement $= 0$ .

Velocity vs. speed. Velocity is signed (a direction); speed is its magnitude. There is no negative speed, just as there is no negative distance.

Average velocity: $v_{\text{avg}} = \dfrac{\Delta x}{\Delta t}$
Instantaneous velocity: $v = \displaystyle\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$ — the derivative of position.

Acceleration is the rate of change of velocity:

Average acceleration: $a_{\text{avg}} = \dfrac{\Delta v}{\Delta t}$
Instantaneous acceleration: $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$ — the second derivative of position.

Graphical reading of motion. On a position-vs-time graph, velocity is the slope and acceleration is the concavity (concave up → speeding up in the $+$ direction; concave down → accelerating in the $-$ direction). Equivalently, integrating acceleration gives velocity, and integrating velocity gives position — the area under an $a$ - $t$ graph is the change in velocity, and the area under a $v$ - $t$ graph is the displacement.

The kinematic equations (constant acceleration only). With constant $a$ , integrating $a$ once gives velocity and again gives position; the constants of integration turn out to be the initial velocity $v_0$ and initial position $x_0$ . The four standard forms:

\begin{aligned} v &= v_0 + at \\ x &= x_0 + v_0 t + \tfrac{1}{2} a t^{2} \\ x &= x_0 + \tfrac{1}{2}(v_0 + v)\,t \\ v^{2} &= v_0^{2} + 2a\,(x - x_0) \end{aligned}

Pick the equation that contains the three quantities you know plus the one you want — e.g. the last form (no $t$ ) is the one to use when you know velocities and displacement but not time.

Free fall. An object in free fall has constant acceleration directed downward. Taking up as positive, $a = -g$ with $g = 9.8\ \text{m/s}^2$ . All four kinematic equations apply unchanged (just substitute $a = -g$ ). At the highest point of a vertical throw the velocity is momentarily zero (but the acceleration is still $-g$ ).

Worked Example

A ball thrown straight up from a tall building. You throw a ball upward with initial velocity $v_0 = 15\ \text{m/s}$ (up is positive, so $a = -g = -9.8\ \text{m/s}^2$ ), releasing it from $y_0 = 0$ .

Velocity after 1 s — start from the velocity equation, then substitute:

\begin{aligned} v &= v_0 + at \\ &= 15 + (-9.8)(1) = 5.2\ \text{m/s} \end{aligned}

Still positive, so the ball is still rising 1 second in.

Maximum height — at the top $v = 0$ , and no time is given, so use the time-free equation:

\begin{aligned} v^{2} &= v_0^{2} + 2a\,(y - y_0) \\ 0 &= (15)^2 + 2(-9.8)\,\Delta y \\ \Delta y &= \frac{(15)^2}{2(9.8)} \approx 11.5\ \text{m} \end{aligned}

The choice of equation in each part is driven entirely by which quantities are known — step 1 had time and wanted velocity; step 2 had velocities and wanted displacement.

Summary

Position and time are meaningful only relative to a chosen reference frame; pick the convenient one.
Displacement (signed, final − initial) differs from distance; velocity (signed) differs from speed.
Velocity is the derivative of position, acceleration the derivative of velocity — slope and concavity on a position-vs-time graph, respectively.
For constant acceleration, the four kinematic equations apply; choose the one matching your known and unknown quantities.
Free fall is constant acceleration with $a = -g$ ( $g = 9.8\ \text{m/s}^2$ ); at peak height $v = 0$ while $a$ stays $-g$ .