Velocity, Speed, and Acceleration: Study Notes

Velocity, Speed, and Acceleration: Key Concepts

  • Both velocity and acceleration are vectors; they have magnitude (size) and direction.
  • Speed vs. velocity:
    • Speed tells you how fast the distance is changing (path length over time).
    • Velocity tells you how fast the displacement is changing (change in position over time) and includes direction.
  • Displacement vs. distance:
    • Displacement is the straight‑line change in position from the initial to final point.
    • Distance is the total length of the path traveled.
  • Instantaneous vs. average quantities:
    • Average velocity is over a finite time interval:
      \mathbf{v}{avg} = \frac{\Delta \mathbf{r}}{\Delta t} = \frac{\mathbf{r}(tf) - \mathbf{r}(ti)}{tf - t_i}.
    • Instantaneous velocity is the limit as the time interval goes to zero:
      \mathbf{v}(t) = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t}.
  • Acceleration:
    • Acceleration is the rate of change of velocity.
    • Average acceleration is
      \mathbf{a}{avg} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{\mathbf{v}(tf) - \mathbf{v}(ti)}{tf - t_i}.
    • Instantaneous acceleration is
      \mathbf{a}(t) = \lim_{\Delta t \to 0} \frac{\mathbf{v}(t + \Delta t) - \mathbf{v}(t)}{\Delta t}.
  • Sign conventions and intuition:
    • Positive acceleration means velocity is increasing in the positive direction.
    • Negative acceleration means velocity is decreasing (the object is slowing down) in the same direction.
  • Quick intuition:
    • If a car moves east at 30 m/s, its velocity is 30 m/s east; its speed is 30 m/s.
    • If a car has acceleration 8 m/s², its velocity increases by 8 m/s each second (in the direction of motion).
  • Worked context and real‑world relevance:
    • Acceleration values indicate how quickly a vehicle can speed up or slow down, relevant to safety and performance.
    • Units and conversions are essential when solving problems across different unit systems.

Worked Problems and Key Steps

  • General approach:
    • Identify knowns (initial velocity, final velocity, time, acceleration).
    • Decide which formula applies (average velocity/acceleration or kinematic relation v = v0 + a t).
    • Perform unit checks and convert units if needed.
    • Pay attention to sign conventions (positive/negative) for direction.

Problem 1: Velocity‑time table with acceleration

  • Given: initial velocity $v_0 = 15\ \mathrm{m/s}$ and acceleration is described as $8\ \mathrm{m/s^2}$ in text, but the table shows velocity increments of 6 m/s per second, yielding:
    • At $t=0$: $v = 15\ \mathrm{m/s}$
    • At $t=1\ \mathrm{s}$: $v = 21\ \mathrm{m/s}$
    • At $t=2\ \mathrm{s}$: $v = 27\ \mathrm{m/s}$
    • At $t=3\ \mathrm{s}$: $v = 33\ \mathrm{m/s}$
    • At $t=4\ \mathrm{s}$: $v = 39\ \mathrm{m/s}$
    • At $t=5\ \mathrm{s}$: $v = 45\ \mathrm{m/s}$
  • Note on inconsistency:
    • The table corresponds to a constant $a = 6\ \mathrm{m/s^2}$ (since $v = v_0 + a t$). The text later states $a = 8\ \mathrm{m/s^2}$, which would give different values (e.g., $v = 15 + 8t$ giving $23\ \mathrm{m/s}$ at $t=1\ \mathrm{s}$). Be aware of this mismatch when studying.
  • Using the relation $v = v_0 + a t$:
    • If $a = 6\ \mathrm{m/s^2}$: $v(t) = 15 + 6t$; values above.
    • If $a = 8\ \mathrm{m/s^2}$: $v(1) = 23\ \mathrm{m/s}$, etc.
  • Takeaway: acceleration changes velocity by $a$ each second; units are m/s² and velocity units are m/s.

Problem 2: Truck accelerates from 25 km/h to 45 km/h in 40 s

  • Knowns:
    • $vi = 25\ \mathrm{km/h}$, $vf = 45\ \mathrm{km/h}$, $\Delta t = 40\ \mathrm{s}$.
  • Average acceleration (in km/h per second):
    • a{avg} = \frac{vf - v_i}{\Delta t} = \frac{45 - 25}{40} = \frac{20}{40} = 0.5\ \mathrm{km!/h!/s}.
  • To convert to m/s²:
    • $0.5\ \mathrm{(km/h)/s} = 0.5 \times \frac{1000\ \mathrm{m}}{3600\ \mathrm{s}}\ \mathrm{/s} = 0.1389\ \mathrm{m/s^2}.$
    • Approximate value: $a_{avg} \approx 0.139\ \mathrm{m/s^2}$.
  • Notes on unit handling:
    • 1 km/h = (1000 m)/(3600 s) so 1 (km/h)/s = (1000/3600) m/s².
  • Practical takeaway:
    • Even though the problem uses mixed units, you can compute average acceleration in one unit system and convert to another for reporting.

Problem 3: Car accelerates from rest at 3.5 m/s² for 12 s

  • Given: $v_0 = 0$, $a = 3.5\ \mathrm{m/s^2}$, $t = 12\ \mathrm{s}$.
  • Final speed:
    • v = v_0 + a t = 0 + (3.5)(12) = 42\ \mathrm{m/s}.
  • Key idea: starting from rest, velocity increases linearly with time under constant acceleration.

Problem 4: Bus accelerates from 12 m/s at 1.2 m/s² for 15 s

  • Given: $v_i = 12\ \mathrm{m/s}$, $a = 1.2\ \mathrm{m/s^2}$, $t = 15\ \mathrm{s}$.
  • Final speed:
    • vf = vi + a t = 12 + (1.2)(15) = 12 + 18 = 30\ \mathrm{m/s}.
  • Takeaway: linear growth of velocity with time under constant acceleration.

Problem 5: Sporty car braking from 95 mph to rest in 4 s

  • Step 1: Convert initial speed to m/s.
    • Classic conversion used: 1 mph ≈ 0.44704 m/s, or equivalently, 95 mph × (1609.34 m / 1 mile) / (3600 s) ≈ 42.47 m/s.
    • Result: $vi \approx 42.47\ \mathrm{m/s}$; $vf = 0$ (rest).
  • Step 2: Compute average acceleration (constant deceleration):
    • a = \frac{vf - vi}{\Delta t} = \frac{0 - 42.47}{4} \approx -10.6\ \mathrm{m/s^2}.
  • Interpretation of sign:
    • The negative sign indicates a decrease in velocity (braking). Acceleration and velocity are opposite in direction during braking, yielding a negative $a$.
  • Practical note on numbers:
    • Using more precise conversion yields similar result: $a \approx -10.6\ \mathrm{m/s^2}$ (typical braking decelerations for abrupt stops fall in this range).

Significance and connections to foundational principles

  • Key connections:
    • Velocity is the derivative of position with respect to time; acceleration is the derivative of velocity with respect to time. These are the core ideas of kinematics and connect to calculus.
  • Real‑world relevance:
    • Understanding acceleration helps in designing safer vehicles, interpreting braking distances, and analyzing sports performance (e.g., how quickly a car can accelerate from 0 to a target speed).
  • Ethical and practical implications:
    • Accurate calculation of acceleration informs safety standards, speed limits, and braking system design. Misapplication or unit errors can lead to unsafe interpretations of vehicle performance.

Common unit conversions and practical tips

  • Speed units:
    • 1 m/s ≈ 3.6 km/h. Conversely, 1 km/h ≈ 0.2778 m/s.
  • Acceleration units:
    • In SI, acceleration is in m/s².
    • If you have a rate with mixed units like (km/h)/s, convert to m/s² by using 1 (km/h)/s = (1000/3600) m/s².
  • Quick reference: 1 mph ≈ 0.44704 m/s; 1 m/s² is a fairly common unit for acceleration in physics problems.

Summary of key formulas

  • Average velocity: \mathbf{v}_{avg} = \frac{\Delta \mathbf{r}}{\Delta t}
  • Instantaneous velocity: \mathbf{v}(t) = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t+\Delta t) - \mathbf{r}(t)}{\Delta t}
  • Average acceleration: \mathbf{a}_{avg} = \frac{\Delta \mathbf{v}}{\Delta t}
  • Instantaneous acceleration: \mathbf{a}(t) = \lim_{\Delta t \to 0} \frac{\mathbf{v}(t+\Delta t) - \mathbf{v}(t)}{\Delta t}
  • Kinematic relation (constant acceleration): \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t

Practical practice problems to reinforce concepts

  • Try varying initial velocity, acceleration, and time to predict final velocity using $v = v_0 + a t$.
  • Convert between unit systems as needed to compare results (e.g., km/h to m/s, mph to m/s).
  • Check signs: positive acceleration means velocity increases in the positive direction; negative acceleration means velocity decreases (or acceleration in the opposite direction).