Chapter 3 Readings: 3.3 Average and Instantaneous Acceleration

Learning Objectives

• Calculate the average acceleration between two points in time.

• Calculate the instantaneous acceleration given the functional form of velocity.

• Explain the vector nature of instantaneous acceleration and velocity.

• Explain the difference between average acceleration and instantaneous acceleration.

• Find instantaneous acceleration at a specified time on a graph of velocity versus time.

Concept and Practical Importance of Acceleration

• Everyday Context: In casual conversation, to accelerate means to speed up, whereas applying the brake pedal causes a vehicle to slow down. The greater the acceleration, the greater the change in velocity over a given time interval.

• Experimental Physics: In linear particle accelerator experiments, subatomic particles are accelerated to extremely high velocities in collision experiments to reveal fundamental information about subatomic structure and the origin of the universe.

• Astrophysical Acceleration and Cosmic Rays:

  • Cosmic rays are subatomic particles accelerated to extremely high energies in supernovas (exploding massive stars) and active galactic nuclei.

  • Understanding these acceleration processes is vital because cosmic rays consist of highly penetrating radiation capable of damaging electronics aboard spacecraft.

Average Acceleration

• Definition: Average acceleration is defined as the rate at which velocity changes over time.

• Mathematical Definition:   $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_0}{t_f - t_0}$

  • $\bar{a}$ represents the average acceleration (the horizontal bar over $a$ denotes average acceleration).

  • $v$ represents velocity ($v_0$ is initial velocity, $v_f$ is final velocity).

  • $t$ represents time ($t_0$ is initial time, $t_f$ is final time).

• SI Units:

  • Units are meters per second divided by seconds, abbreviated as $\text{m/s}^2$ (meters per second squared, or meters per second per second).

  • This unit explicitly measures by how many meters per second the velocity changes during each second of time.

• Vector Properties:

  • Velocity is a vector quantity possessing both magnitude (speed) and direction.

  • A change in velocity $\Delta v$ can result from a change in magnitude (speeding up or slowing down), a change in direction, or both.

  • Directional Example: If a runner traveling at $10\,\text{km/h}$ due east slows to a stop, reverses direction, and continues running at $10\,\text{km/h}$ due west, her velocity has changed significantly as a result of the change in direction, even though the speed magnitude at the ends of the run is identical.

  • Acceleration is a vector in the exact same direction as the change in velocity vector $\Delta v$.

• Direction of Motion vs. Direction of Acceleration:

  • Acceleration is in the direction of the change in velocity, but it is not always in the direction of motion.

  • When an object slows down, its acceleration points in the direction opposite to its motion.

  • Terminology Note: While slowing down is colloquially called "deceleration" (such as a subway train in São Paulo, Brazil accelerating opposite to its motion as it enters a station), the term deceleration is avoided in formal physical analysis because it is not a vector and does not point in a specific direction relative to a coordinate frame.

  • In formal analysis, vector signs (+ or -) are used within a designated coordinate system. An object slowing down while moving in a positive direction undergoes negative acceleration.

• Motion under Constant Negative Acceleration:

  • If an object has a positive velocity relative to a chosen origin and experiences a constant negative acceleration, it slows down, comes to rest, reverses direction, and eventually passes through the origin traveling in the opposite direction.

Example Problems: Average Acceleration

Example 3.5: Calculating Average Acceleration — A Racehorse Leaves the Gate

• Problem Statement: A racehorse coming out of the gate accelerates from rest to a velocity of $15.0\,\text{m/s}$ due west in $1.80\,\text{s}$. What is its average acceleration?

• Strategy & Coordinate Setup:

  • Define a one-dimensional coordinate system where East is positive (+) and West is negative (-).

  • Westbound velocity is assigned a negative sign.

• Known Variables:

  • Initial velocity: $v_0 = 0\,\text{m/s}$

  • Final velocity: $v_f = -15.0\,\text{m/s}$

  • Elapsed time: $\Delta t = 1.80\,\text{s}$

• Step-by-Step Calculation:

  1. Calculate change in velocity:      $\Delta v = v_f - v_0 = -15.0\,\text{m/s} - 0\,\text{m/s} = -15.0\,\text{m/s}$

  2. Compute average acceleration:      $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{-15.0\,\text{m/s}}{1.80\,\text{s}} = -8.33\,\text{m/s}^2$

• Significance:

  • The negative sign indicates acceleration is directed toward the west.

  • An acceleration of $8.33\,\text{m/s}^2$ due west means the horse increases its velocity by $8.33\,\text{m/s}$ due west every second.

  • This value represents average acceleration because the horse's motion is non-uniform. An acceleration of this magnitude requires the rider to hold on with a force nearly equal to their own weight.

Check Your Understanding 3.3

• Problem Statement: Protons in a linear accelerator are accelerated from rest to $2.0 \times 10^7\,\text{m/s}$ in $10^{-4}\,\text{s}$. What is the average acceleration of the protons?

• Calculation:   $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{2.0 \times 10^7\,\text{m/s} - 0\,\text{m/s}}{10^{-4}\,\text{s}} = 2.0 \times 10^{11}\,\text{m/s}^2$

Instantaneous Acceleration

• Definition: Instantaneous acceleration $a$ is the acceleration of an object at a specific instant in time.

• Calculus Formulation:

  • Obtained by taking the average acceleration over a time interval $\Delta t$ and letting $\Delta t \rightarrow 0$:     $a(t) = \frac{d}{dt} v(t)$

  • Instantaneous acceleration is the first derivative of the velocity function $v(t)$ with respect to time.

• Graphical Interpretation:

  • On a graph of velocity versus time $v(t)$, instantaneous acceleration at time $t$ is the slope of the tangent line to the velocity curve at that time.

  • As $\Delta t \rightarrow 0$, the average acceleration $\bar{a} = \frac{\Delta v}{\Delta t}$ approaches instantaneous acceleration.

• Velocity Maxima and Minima:

  • At points where velocity reaches a maximum or minimum value, the slope of the tangent line is zero ($\frac{dv}{dt} = 0$).

  • Thus, the zeros (roots) of the instantaneous acceleration function $a(t) = 0$ correspond to local maximum or minimum points of the velocity function.

  • Linear velocity-versus-time graphs with constant negative slope yield a constant negative acceleration.

Example Problems: Instantaneous Acceleration

Example 3.6: Calculating Instantaneous Acceleration

• Problem Statement: A particle is in motion and accelerating. The functional form of velocity is:   $v(t) = (20\,\text{m/s}^2)t - (5\,\text{m/s}^3)t^2$

  • a. Find the functional form of acceleration.

  • b. Find the instantaneous velocity at $t = 1\,\text{s}$, $2\,\text{s}$, $3\,\text{s}$, and $5\,\text{s}$.

  • c. Find the instantaneous acceleration at $t = 1\,\text{s}$, $2\,\text{s}$, $3\,\text{s}$, and $5\,\text{s}$.

  • d. Interpret the results of (c) in terms of the directions of the acceleration and velocity vectors.

• Solution:

  • Part a (Functional form of acceleration):     $a(t) = \frac{dv(t)}{dt} = (20\,\text{m/s}^2) - (10\,\text{m/s}^3)t$

  • Part b (Instantaneous velocity values):

  • $v(1\,\text{s}) = (20)(1) - (5)(1)^2 = 15\,\text{m/s}$

  • $v(2\,\text{s}) = (20)(2) - (5)(2)^2 = 20\,\text{m/s}$

  • $v(3\,\text{s}) = (20)(3) - (5)(3)^2 = 15\,\text{m/s}$

  • $v(5\,\text{s}) = (20)(5) - (5)(5)^2 = -25\,\text{m/s}$

  • Part c (Instantaneous acceleration values):

  • $a(1\,\text{s}) = 20 - 10(1) = 10\,\text{m/s}^2$

  • $a(2\,\text{s}) = 20 - 10(2) = 0\,\text{m/s}^2$

  • $a(3\,\text{s}) = 20 - 10(3) = -10\,\text{m/s}^2$

  • $a(5\,\text{s}) = 20 - 10(5) = -30\,\text{m/s}^2$

  • Part d (Physical Interpretation):

  • At $t = 1\,\text{s}$: Velocity $v(1\,\text{s}) = 15\,\text{m/s}$ is positive and acceleration $a(1\,\text{s}) = 10\,\text{m/s}^2$ is positive. Because both vectors point in the same direction, the particle is moving faster (speeding up).

  • At $t = 2\,\text{s}$: Velocity increases to its maximum value of $20\,\text{m/s}$, where acceleration is zero. Maximum velocity occurs where the slope of velocity is zero ($a(t) = 0$).

  • At $t = 3\,\text{s}$: Velocity $v(3\,\text{s}) = 15\,\text{m/s}$ is positive, but acceleration $a(3\,\text{s}) = -10\,\text{m/s}^2$ is negative. Acting in opposite directions, acceleration causes the particle to slow down.

  • At $t = 5\,\text{s}$: Velocity $v(5\,\text{s}) = -25\,\text{m/s}$ is negative and acceleration $a(5\,\text{s}) = -30\,\text{m/s}^2$ is increasingly negative. Between $t = 3\,\text{s}$ and $t = 5\,\text{s}$, velocity reduced to zero and reversed direction. Both vectors now point in the negative direction, so the particle is speeding up in the opposite direction.

• Significance & Real-World Analogy:

  • Numerical analysis complements graphical analysis to provide a complete view of motion.

  • Real-world physical example: A car increasing velocity to a maximum, slowing down to a temporary stop, and then reversing direction while accelerating backwards.

Check Your Understanding 3.4

• Problem Statement: An airplane lands on a runway traveling east. Describe its acceleration.

• Explanation:

  • The airplane has a positive velocity directed eastward.

  • As it lands and slows down, its change in velocity vector points westward.

  • Consequently, its acceleration is directed west (negative acceleration in an eastward-positive coordinate system).

Orders of Magnitude of Acceleration

• Scale Independence: Acceleration magnitudes range over many orders of magnitude across natural and technological systems and are independent of an object's size or mass.

• Typical Values of Acceleration (Table 3.2):

• Temporal Variability of Acceleration:

  • Acceleration can change significantly over time during motion (e.g., a drag racer experiences huge acceleration immediately after launch, tapering off as constant speed is reached).

  • For motions with extreme acceleration fluctuations (e.g., a package on a post office conveyor belt bumping forward and backward), analysis requires dividing time into short intervals (such as $0\text{--}1.0\,\text{s}$) over which acceleration can be treated as constant or nearly constant.

Interactive Simulation

• Moving Man Simulation: An interactive tool designed to explore position, velocity, and acceleration graphs by manually moving a figure back and forth or setting position, velocity, and acceleration values to automate and plot motion graphs.