Motion Along a Straight Line
Position, velocity, and acceleration — the grammar of one-dimensional motion
- How to pin down an object's position and its displacement on a straight axis.
- The difference between average velocity, average speed, and the instantaneous versions.
- That velocity is the slope of \(x(t)\) and acceleration the slope of \(v(t)\).
- The five constant-acceleration equations and when they apply.
- How free fall is just constant acceleration with \(a = -g\).
What Is Physics?
A big part of physics is the study of motion — how fast things move and how far they travel in a given time. Race engineers tune cars with it, geologists track tectonic plates with it, and doctors map blood flow with it. We start with the simplest case: an object moving back and forth along a single straight line. This is called one-dimensional motion.
Motion: The Ground Rules
Describing and comparing motion is the branch of physics called kinematics. To keep this chapter simple, we restrict ourselves in three ways:
- Straight line only. The path may be horizontal, vertical, or slanted — but it must be straight.
- Forces come later. We describe the motion (speeding up, slowing, reversing) without yet asking what causes it. Forces wait until Chapter 5.
- Treat the object as a particle. A point-like object, or one whose every part moves the same way — like a sled, not a tumbling tumbleweed.
Position & Displacement
We locate an object by its position on an axis — say the \(x\) axis — measured from the origin (the zero point). Positions to the right are positive; to the left, negative. A coordinate of \(x = +5 m\) sits 5 m right of the origin; \(x = -5 m\) is the same distance to the left.
A change in position is the displacement:
Displacement depends only on the start and end points. Walk from \(x = 5 m\) out to 200 m and back to 5 m, and your displacement is zero — even though you covered 390 m. Displacement is a vector: it carries both a size (magnitude) and a direction (the + or − sign).
Average Velocity & Average Speed
The clearest way to picture motion is a graph of position versus time, \(x(t)\). The average velocity over an interval is the displacement divided by the time taken — which is exactly the slope of the line joining the two points on that graph:
Average speed is different. It ignores direction and uses the total distance covered, so it never carries a sign:
Instantaneous Velocity & Speed
Usually "how fast" means right now, not averaged over an interval. Shrink the time interval toward zero and the average velocity approaches the instantaneous velocity — the derivative of position:
Acceleration
When velocity changes, the object accelerates. Average acceleration is the change in velocity over the time interval; instantaneous acceleration is the derivative of velocity — and therefore the second derivative of position:
The sign of \(a\) gives its direction on the axis, not whether the object speeds up. The rule: if velocity and acceleration have the same sign, speed increases; if opposite signs, speed decreases. A car braking from −25 m/s has a positive acceleration while slowing down.
The Constant-Acceleration Equations
A great many situations — a car pulling away from a light, an object in free fall — involve (very nearly) constant acceleration. For those cases, five handy equations link the five quantities: displacement \(x - x_{0}\), initial velocity \(v_{0}\), final velocity \(v\), acceleration \(a\), and time \(t\). Each equation leaves one of them out.
| Equation | Missing quantity |
|---|---|
| v = v0 + at | x − x0 |
| x − x0 = v0t + ½at2 | v |
| v2 = v02 + 2a(x − x0) | t |
| x − x0 = ½(v0 + v)t | a |
| x − x0 = vt − ½at2 | v0 |
These five equations are valid only for constant acceleration. If \(a\) changes with time, fall back on calculus: \(v = \int a\,dt\) and \(x = \int v\,dt\).
Free-Fall Acceleration
Drop the air resistance, and every object near Earth's surface falls with the same constant downward acceleration — regardless of mass, shape, or density. A feather and an apple fall together in a vacuum. That rate is the free-fall acceleration, magnitude \(g\).
Use the Table 2-1 equations, but switch to a vertical \(y\) axis (positive up) and set \(a = -g\). The acceleration is the same on the way up and the way down — including the instant at the very top, where the velocity is momentarily zero.
Graphical Integration
Because \(a = dv/dt\) and \(v = dx/dt\), we can run the derivatives backwards using areas under graphs:
Slope of \(x(t)\) → velocity. Slope of \(v(t)\) → acceleration.
Area under \(a(t)\) → change in velocity. Area under \(v(t)\) → change in position.
Putting It to Work
Problem. You drive 8.4 km at 70 km/h, run out of fuel, then walk 2.0 km in 30 min to a station. Find your overall displacement, your average velocity, and your average speed for the drive-plus-walk.
Displacement. Taking the start as \(x = 0\), the station is at \(8.4 + 2.0 = 10.4 \mathrm{km}\), so \(\Delta x = +10.4 \mathrm{km}\).
Times. Driving takes \(t = 8.4/70 = 0.12 h\); walking takes 0.50 h; total \(\Delta t = 0.62 h\).
The average speed here is the same (no doubling back): \(10.4 \mathrm{km} / 0.62 h \approx 17 \mathrm{km}/h\). Add a return walk and they would diverge.
Problem. A particle moves as \(x = 4 - 27t + t^{3}\) (metres, seconds). Find \(v(t)\) and \(a(t)\), and the time when \(v = 0\).
Set \(v = 0\): \(3t^{2} = 27 \to t = \pm 3 s\). The particle is momentarily at rest 3 s before and 3 s after the clock reads zero. Just before \(t = 3 s\) the velocity is negative while the acceleration is positive — opposite signs — so it is slowing to that stop, then reverses.
Problem. A ball is thrown straight up at \(v_{0} = 12 m/s\). How long to reach the top, and how high does it rise? (Take \(a = -g = -9.8 m/s^{2}\).)
Time to top (where \(v = 0\)), from \(v = v_{0} + at\):
It climbs about 7.3 m and takes 1.2 s to get there — and the same 1.2 s to come back down.
Chapter Summary
Position locates an object on an axis; displacement \(\Delta x = x_{2} - x_{1}\) is a vector (final minus initial), not total distance.
\(v_{\text{avg}} = \Delta x/\Delta t\) (slope of x–t); \(v = dx/dt\). Speed is |v|; average speed uses total distance.
\(a = dv/dt = d^{2}x/dt^{2}\). Same sign as v → speeding up; opposite → slowing.
Five linked equations (Table 2-1). Pick the one missing the quantity you don't have and don't need.
Constant acceleration with \(a = -g \approx -9.8 m/s^{2}\) on a vertical axis, positive up.
Slopes give derivatives (v, a); areas give integrals (Δv, Δx).
Problems
Sketch an \(x(t)\) or \(v(t)\) graph where it helps, and watch your signs.
- During a hard sneeze your eyes shut for 0.50 s. At 90 km/h, how far does your car travel in that time?
- A car covers 40 km at 30 km/h, then another 40 km at 60 km/h in the same direction. Find (a) the average velocity and (b) the average speed for the full 80 km.
- A car goes up a hill at a constant 40 km/h and back down at 60 km/h. What is the average speed for the round trip? (It is not 50 km/h.)
- A particle's position is \(x = 4 - 12t + 3t^{2}\) (m, s). Find its velocity at \(t = 1 s\), and say whether it is then moving in the +x or −x direction.
- At one instant a particle moves at 18 m/s in the +x direction; 2.4 s later it moves at 30 m/s in the −x direction. Find its average acceleration.
- An electric car accelerates from rest at 2.0 m/s² up to 20 m/s, then brakes at 1.0 m/s² to a stop. Find (a) the total time and (b) the total distance.
- On a dry road a car brakes at a constant 4.92 m/s². Starting at 24.6 m/s, (a) how long to stop and (b) how far does it travel?
- With what speed must a ball be thrown straight up to reach a maximum height of 50 m, and (b) how long is it in the air?
- A stone is thrown straight down at 12.0 m/s from a 30.0 m rooftop. (a) How long to reach the ground and (b) at what speed does it land?
- A hot-air balloon rises at 12 m/s; at 80 m up, a package is released. (a) How long until it lands and (b) at what speed? (Mind that the package starts with the balloon's upward velocity.)