Motion at a Constant Speed: Distances, Times, and Powers of Ten

Motion at a Constant Speed

Introduction

This section introduces the fundamental concepts of motion at a constant speed, focusing on the relationships between distance, time, and speed. It also covers unit conversions and the use of scientific notation for large and small quantities, which are essential for solving physics problems efficiently.

Distances, Times, and Powers of Ten

Scientific Notation and Unit Conversions

• Scientific notation is used to express very large or very small numbers in a compact form, such as $1 \text{ km} = 10^3 \text{ m}$.

• Common unit conversions include:

  • $1 \text{ m} = 100 \text{ cm} = 10^3 \text{ mm} = 10^6 \mu\text{m} = 10^9 \text{ nm}$

  • $1 \text{ km} = 1,000 \text{ m}$

  • $1 \text{ mi} = 1.609 \text{ km} = 1,609 \text{ m}$

  • $1 \text{ foot} = 30.5 \text{ cm} = 0.305 \text{ m}$

  • $1 \text{ h} = 60 \text{ min}$

  • $1 \text{ year} \approx 3.15 \times 10^7 \text{ s}$ (Additional info: standard value for seconds in a year)

• Powers of ten help compare and calculate with large numbers, such as population or national debt.

Example: U.S. National Debt Calculation

• Given: U.S. national debt = $37.2 \times 10^{12}$ dollars, U.S. population = $340 \times 10^6$ people.

• Debt per person: $\frac{37.2 \times 10^{12}}{340 \times 10^6} \approx 1.1 \times 10^5$ dollars.

• Counting the debt in $100 bills at one bill per second would take $1.18 \times 10^4$ years.

Example: Circumference of the Earth and Population

• Earth's equator length: $40,000 \text{ km} = 4 \times 10^7 \text{ m}$.

• If 8 billion people ($8 \times 10^9$) stood 2 m apart, the line would be $1.6 \times 10^{10}$ m long.

• This line would circle the equator $\frac{1.6 \times 10^{10}}{4 \times 10^7} = 400$ times.

Speed, Velocity, and Unit Conversion

Definitions and Formulas

• Speed is the distance traveled per unit time: $v = \frac{l}{t}$.

• Velocity is speed with a specified direction.

• Average speed is total distance divided by total time: $\bar{v} = \frac{l_{\text{tot}}}{t_{\text{tot}}}$.

• Instantaneous speed is the speed at a specific moment, as shown by a speedometer.

Unit Conversion Examples

• Convert $55.0 \text{ mi/h}$ to $\text{km/h}$ and $\text{m/s}$:

  • $55.0 \times 1.609 \text{ km/h} \approx 88.5 \text{ km/h}$

  • $88.5 \times 1000 \text{ m} / 3600 \text{ s} = 24.6 \text{ m/s}$

• Usain Bolt's 100-m world record ($9.5805$ s):

  • Speed: $\frac{100 \text{ m}}{9.5805 \text{ s}} = 10.44 \text{ m/s}$

  • Convert to mi/h: $10.44 \times (1 \text{ mi}/1609 \text{ m}) \times (3600 \text{ s}/1 \text{ h}) \approx 23.4 \text{ mi/h}$

Solving Problems Involving Average Speed

Key Steps

• Identify total distance and total time.

• For multiple segments at different speeds, calculate time for each segment, sum times, and divide total distance by total time.

Example Problems

• Problem 5: Distance from New York to Boston is 215 mi at 55 mi/h.

  • Time: $t = \frac{l}{v} = \frac{215}{55} \approx 3.9 \text{ h}$

• Problem 6: Student walks 60 m, stops for 1 min halfway.

  • Total time: $t_{\text{tot}} = 60 \text{ s} + 60 \text{ s} + 20 \text{ s} = 100 \text{ s}$

  • Average speed: $\bar{v} = \frac{60 \text{ m}}{100 \text{ s}} = 0.6 \text{ m/s}$

• Problem 7: Car travels 150 mi: first half at 45 mi/h, second half at 75 mi/h.

  • Time for each half: $t_1 = \frac{75}{45}$ h, $t_2 = \frac{75}{75}$ h

  • Total time: $t_{\text{tot}} = 1.67 + 1 = 2.7$ h

  • Average speed: $\bar{v} = \frac{150}{2.7} \approx 56 \text{ mi/h}$

Graphical Representation of Motion

Position vs. Time Graphs

• For motion at constant speed along the x-axis: $x = x_0 + vt$

• Graph shows straight lines for constant velocity, with slope equal to velocity.

• Negative slope indicates motion in the opposite direction.

Example: Average Speed from a Graph

• If each block is 100 m and 9 blocks are covered in 26 min:

• Total distance: $d = 9 \times 100 = 900 \text{ m}$

• Total time: $t = 26 \text{ min} = 26 \times (1 \text{ h}/60) = 0.433 \text{ h}$

• Average speed: $\bar{v} = \frac{900}{26} \times \frac{1}{1000} \times 60 \approx 2.1 \text{ km/h}$

Oscillatory Motion: The Pendulum

Pendulum Period and Average Speed

• The period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$

• For $l = 2 \text{ m}$, $g = 9.8 \text{ m/s}^2$:

• During a 90° swing, the sphere travels a quarter of a circle of radius $l$.

• Distance traveled in one period: $\pi l$

• Average speed: $\bar{v} = \frac{\pi l}{2\pi \sqrt{l/g}} = \frac{1}{2} \sqrt{g l}$

• For $l = 2 \text{ m}$: $\bar{v} = 0.5 \sqrt{9.8 \times 2} \approx 2.2 \text{ m/s}$

Calculus in Kinematics (Optional)

Derivatives and Integrals

• Instantaneous velocity is the derivative of position with respect to time:

$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

• Distance as an integral of velocity over time:

$x = \int v(t) dt$

• For constant velocity $v_0$:

$x = \int v_0 dt = v_0 t$

Additional info: Calculus-based definitions are foundational for advanced kinematics and will be explored in more detail in later chapters.

Summary Table: Key Quantities and Formulas

Quantity	Symbol	Formula	SI Unit
Distance	l, d, x	-	meter (m)
Time	t	-	second (s)
Speed	v	$v = \frac{l}{t}$	m/s
Average Speed	\bar{v}	$\bar{v} = \frac{l_{\text{tot}}}{t_{\text{tot}}}$	m/s
Instantaneous Speed	v	$v = \frac{dx}{dt}$	m/s
Pendulum Period	T	$T = 2\pi \sqrt{\frac{l}{g}}$	s