Chapter 2 Notes: Motion Along a Straight Line

2-1 Position, Displacement, and Average Velocity

Physics focus: motion of objects moving along a single axis; treat such objects as point-like particles if all parts move in the same direction and rate.
Position along an axis: the location x read on a scaled axis (e.g., x-axis).
Sign convention on an axis:
- Positive direction: increasing coordinates (to the right in the figure).
- Negative direction: decreasing coordinates (to the left).
Displacement Δx:
- Definition: change in position when moving from x1 to x2 in time interval Δt = t2 − t1.
- Vector quantity with sign; Δx > 0 if motion is in the positive direction, Δx < 0 if in the negative direction.
- Formula: $\Delta x = x2 - x1$
- Example: moving from x1 = 5 m to x2 = 12 m → $\Delta x = 12\,\text{m} - 5\,\text{m} = +7\,\text{m}$ ; from x1 = 5 m to x2 = 1 m → $\Delta x = 1\,\text{m} - 5\,\text{m} = -4\,\text{m}$
- The final displacement may be zero even if the path covers distance (e.g., 5 m → 200 m → 5 m gives Δx = 0).
- Displacement is a vector quantity with magnitude and direction; magnitude is the straight-line distance between initial and final positions.
Average velocity v_avg over interval Δt:
- Definition: ratio of displacement to time interval.
- Formula: $v{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x2 - x1}{t2 - t_1}$
- Sign of v_avg matches the direction of motion (positive if moving in +x, negative if in -x).
- On a graph of x versus t, v_avg is the slope of the straight line joining the points (x1, t1) and (x2, t2).
Average speed s_avg over Δt:
- Definition: total distance traveled divided by the time interval.
- Formula: $\text{s}_{\text{avg}} = \frac{\text{distance traveled in the interval}}{\Delta t}$
- Unlike v_avg, savg has no sign (it is a nonnegative quantity).
Relationship to everyday physics: study of how fast and how far objects move; real-world relevance spans NASCAR, geology, medicine, and transportation safety.
One-dimensional motion: motion along a straight line (vertical, horizontal, or slanted).
Checkpoint 1 (quick reasoning): Given three initial/final positions, determine which yield a negative displacement.
- Pairs: (a) −3 m → +5 m; (b) −3 m → −7 m; (c) 7 m → −3 m
- Answers: (b) Δx = (−7) − (−3) = −4 m; (c) Δx = (−3) − 7 = −10 m; thus (b) and (c) have negative displacement.

2-2 Instantaneous Velocity and Speed

Instantaneous velocity v at time t: rate at which position x changes at that exact time.
Definition via calculus: v(t) = \frac{dx}{dt} (the slope of the x(t) curve at time t).
Speed: the magnitude of velocity; $\text{speed} = |v|$ ; speed has no direction.
Instantaneous velocity vs average velocity:
- Average velocity is over a finite interval; instantaneous velocity is limit as the interval shrinks to zero.
- Graphically, on an x vs t plot, the slope at a single time t gives v(t).
Key relation: velocity is a vector quantity and has both magnitude and direction.
Example intuition: a stationary object has v = 0 → slope of x(t) is zero at that time.
Elevator-cab illustration (conceptual): x(t) curve for motion along an axis; slope of x(t) yields v(t); slope of v(t) yields a(t) (see Fig. 2-6 in the module).

2-3 Acceleration

Average acceleration a_avg over a time interval Δt:
- Definition: change in velocity divided by the time over which the change occurs.
- Formula: $a{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v2 - v1}{t2 - t_1}$
- Sign indicates the acceleration's direction along the axis.
Instantaneous acceleration a(t):
- Definition: derivative of velocity with respect to time, or second derivative of position.
- Formulas: $a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}$
On a graph of velocity versus time (v vs t), the acceleration at time t is the slope of the v(t) curve at that time.
Units: $\text{[acceleration]} = \text{m s}^{-2}$ (often written as m/s^2).
Vector nature of acceleration: has magnitude and direction; positive a means acceleration in +x, negative a in −x.
Elevator example (qualitative):
- When v is constant, a = 0.
- During acceleration, v increases in the positive direction (a > 0) if moving in +x.
- When decelerating, magnitude of a may be larger even if time to stop is shorter (steeper slope on v(t) during braking).
Practical interpretation: signs indicate direction; if velocity and acceleration have the same sign, speed increases; if opposite signs, speed decreases.
Common cues: large accelerations (e.g., up to 3g on roller coasters) can be described in g-units, where g ≈ 9.8 m/s^2.

Quick checkpoints and concepts

Checkpoint 2: Given four position-time equations, identify where velocity is constant and where velocity is negative.
- Equations (examples):
- (1) $x = 3t - 2$ → v(t) = 3 (constant, positive)
- (2) $x = -4t^2 - 2$ → v(t) = -8t (not constant)
- (3) $x = \frac{2}{t^2}$ → v(t) = -\frac{4}{t^3} (not constant)
- (4) $x = -2$ → v(t) = 0 (constant, zero velocity)
- Negative velocity occurs for (2) at t > 0 and for (3) at t > 0.
Checkpoint 3: Interpreting velocity and acceleration signs: if velocity and acceleration have the same sign, speed increases; if opposite signs, speed decreases.

Sample Problems

Sample Problem 2.01: Average velocity, beat-up pickup truck
- The journey: 8.4 km at 70 km/h, then 2.0 km walking to a gas station.
- (a) Displacement from start to station:
- Treat first position x1 = 0 and final x2 = 8.4 km + 2.0 km = 10.4 km.
- $\Delta x = x2 - x1 = 10.4\ \text{km}$
- Answer: $\Delta x = 10.4\ \text{km}$ (positive direction).
- (b) Time interval for driving to the station:
- Driving displacement Δxdr = 8.4 km; vavg,dr = 70 km/h.
- $a) \Delta t{dr} = \frac{\Delta x{dr}}{v_{avg,dr}} = \frac{8.4\ \text{km}}{70\ \text{km/h}} = 0.12\ \text{h}$
- (c) Average velocity v_avg for the entire trip (start to station):
- Total displacement Δxtotal = 10.4 km; total time Δttotal = 0.12 h + 0.50 h = 0.62 h.
- $v{avg} = \frac{\Delta x{total}}{\Delta t_{total}} = \frac{10.4\ \text{km}}{0.62\ \text{h}} \approx 16.8\ \text{km/h}$
- Graphical check: slope from origin to the Station on the x(t) plot gives the same value, $v_{avg} = 16.8\ \text{km/h}$ .
- (d) If you spend 45 minutes pumping gas and walking back to the truck, find the average speed for the entire trip (start to return):
- Total distance: 8.4 + 2.0 + 2.0 = 12.4 km.
- Total time: 0.12 h + 0.50 h + 0.75 h = 1.37 h.
- $\text{s}_{avg} = \frac{\text{distance}}{\text{time}} = \frac{12.4\ \text{km}}{1.37\ \text{h}} \approx 9.0\ \text{km/h}$
Sample Problem 2.02: Velocity and slope of x versus t, elevator cab
- Given x(t) for an elevator cab, plot v(t) as the slope of x(t).
- Key takeaway: velocity is the slope of the x(t) curve; a(t) is the slope of the v(t) curve (i.e., the derivative of velocity).
Sample Problem 2.03: Acceleration and dv/dt
- Provided an x(t) on the x-axis; (a) find v(t) = dx/dt and a(t) = dv/dt.
- (b) Determine if/when v = 0.
- (c) Describe the motion for t ≥ 0 using x(t), v(t), a(t).
- General method: differentiate x(t) to get v(t); differentiate v(t) to get a(t); analyze signs and zeros to understand motion.

How to connect the concepts

From position to velocity:
- v(t) is the slope of x(t) at time t: $v(t) = \frac{dx}{dt}$ .
From velocity to acceleration:
- a(t) is the slope of v(t) at time t, or the second derivative of position: $a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}$ .
From velocity to displacement:
- Displacement over a time interval is the integral of velocity: approximately the area under the v(t) curve between t1 and t2; for constant v, Δx = v × Δt; for variable v, Δx = ∫_{t1}^{t2} v(t) dt.

Foundational notes and signs

Sign conventions:
- Positive velocity or acceleration indicates motion or change in the positive x direction.
- If velocity and acceleration share the same sign, speed increases; if signs are opposite, speed decreases.
Units recap:
- Position: meters (m)
- Velocity: meters per second (m/s)
- Acceleration: meters per second squared (m/s^2)
Practical context: a few examples show why these concepts matter in real life, from race cars to earthquakes to medical diagnostics and everyday driving.

Summary of key formulas (LaTeX)

Displacement: $\Delta x = x2 - x1$
Average velocity: $v{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x2 - x1}{t2 - t_1}$
Average speed: $\text{s}_{\text{avg}} = \frac{\text{distance traveled}}{\Delta t}$
Instantaneous velocity: $v(t) = \frac{dx}{dt}$
Acceleration (average): $a{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v2 - v1}{t2 - t_1}$
Acceleration (instantaneous) and position relations: $a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}$
Sign interpretation rule: if sgn(v) = sgn(a) then speed is increasing; if sgn(v) ≠ sgn(a) then speed is decreasing.