#Distance & Speed as Integrals

#Table of Contents

1. Introduction to Distance and Speed
2. Distance vs. Displacement and Speed vs. Velocity
3. Finding Distance Using Integration
4. Impact of Speed vs. Velocity in Calculations
5. Worked Example
6. Practice Questions
7. Glossary
8. Summary and Key Takeaways

#Introduction to Distance and Speed

Understanding the concepts of distance and speed is fundamental for solving problems in physics and calculus. These concepts are distinct from displacement and velocity, which are often confused by students. This guide will clarify these differences and explain how to use integration to find distances.

#Distance vs. Displacement and Speed vs. Velocity

#Definitions and Differences

• Distance refers to the total length of the path traveled by an object, regardless of direction. It is a scalar quantity and always positive.
• Displacement is a vector quantity that measures the change in position of an object. It has both magnitude and direction.
• Speed is the magnitude of velocity. It is a scalar quantity and always positive.
• Velocity is a vector quantity that describes the rate of change of position. It has both magnitude and direction.

**Example:** - Traveling **3 meters forwards** and **3 meters backwards** results in a **distance** of **6 meters**. - The **displacement** after this motion is **zero**. - Traveling at **10 meters per second forward** is a velocity of **+10 meters per second**. - Traveling at **10 meters per second backwards** is a velocity of **-10 meters per second**. - In both cases, the **speed** is **10 meters per second**.

#Finding Distance Using Integration

To find the total distance traveled by an object between two times, we use the integral of the absolute value of the velocity function.

#Formula

$$\text{Distance} = \int_{t_1}^{t_2} |v(t)| , dt$$

#Example Calculation

Consider a velocity function $v = 2t - 6$.

1. To find the change in displacement between $t=0$ and $t=6$, calculate: $$\int_{0}^{6} (2t - 6) , dt$$ This integral evaluates to zero because: $$\int_{0}^{3} (2t - 6) , dt = -9 \quad \text{and} \quad \int_{3}^{6} (2t - 6) , dt = +9$$

2. To find the distance traveled between $t=0$ and $t=6$, calculate: $$\int_{0}^{6} |2t - 6| , dt$$ This can be split into two parts: $$\int_{0}^{3} |2t - 6| , dt + \int_{3}^{6} |2t - 6| , dt$$ Since the integral from 0 to 3 is negative, we take the absolute value: $$| \int_{0}^{3} (2t - 6) , dt | + \int_{3}^{6} (2t - 6) , dt = 9 + 9 = 18$$

#Impact of Speed vs. Velocity in Calculations

#Key Points

• Pay attention to whether the question refers to speed or velocity.
• When velocity changes sign, it impacts the calculation of distance vs. displacement.

**Example:** - A particle with **velocity 5 meters per second** decreases its velocity by **20 meters per second**: - New velocity: **-15 meters per second** - New speed: **15 meters per second** - A particle with **velocity -8 meters per second** increases its velocity by **10 meters per second**: - New velocity: **2 meters per second** - Speed has decreased from **8 meters per second** to **2 meters per second**.

#Worked Example

#Problem Statement

The acceleration of a particle over the interval $0 \le t \le 6\pi$ is given by: $$a(t) = \frac{3}{2} \cos \left(\frac{t}{2}\right)$$ where $v$ is measured in feet per second and $t$ in seconds. At $t=0$, the particle is at rest.

#(a) Calculate the change in speed between $t=\frac{3\pi}{2}$ and $t=3\pi$. State if this is an increase or decrease.

Solution:

1. Integrate the acceleration function to find the velocity function: $$v(t) = \int a(t) , dt = \int \frac{3}{2} \cos \left(\frac{t}{2}\right) , dt = 3 \sin\left(\frac{t}{2}\right) + C$$

2. Given that $v(0) = 0$: $$0 = 3 \sin(0) + C \implies C = 0$$

3. Thus, the velocity function is: $$v(t) = 3 \sin\left(\frac{t}{2}\right)$$

4. Find velocities at $t=\frac{3\pi}{2}$ and $t=3\pi$: $$v\left(\frac{3\pi}{2}\right) = 3 \sin\left(\frac{3\pi}{4}\right) \approx 2.121$$ $$v\left(3\pi\right) = 3 \sin\left(\frac{3\pi}{2}\right) = -3$$

5. Convert these to speeds:

• Speed at $t=\frac{3\pi}{2}$ is approximately $2.121$ feet per second.
• Speed at $t=3\pi$ is $3$ feet per second.

6. The speed has increased by: $$3 - 2.121 = 0.879 \text{ feet per second}$$

#(b) Find the total distance traveled by the particle between $t=\pi$ and $t=4\pi$.

Solution:

1. The distance traveled is: $$\int_{\pi}^{4\pi} |v(t)| , dt = \int_{\pi}^{4\pi} |3 \sin \left(\frac{t}{2}\right)| , dt$$

2. Split the integral at points where the velocity function changes sign: $$\int_{\pi}^{2\pi} 3 \sin \left(\frac{t}{2}\right) , dt + \left| \int_{2\pi}^{4\pi} 3 \sin \left(\frac{t}{2}\right) , dt \right|$$

3. Calculate each integral: $$\int_{\pi}^{2\pi} 3 \sin \left(\frac{t}{2}\right) , dt = 6$$ $$\left| \int_{2\pi}^{4\pi} 3 \sin \left(\frac{t}{2}\right) , dt \right| = 12$$

4. Total distance: $$6 + 12 = 18 \text{ feet}$$

#Practice Questions

Practice Question

1. Compute the total distance traveled by a particle with velocity $v(t) = t^2 - 4t + 3$ over the interval $0 \le t \le 5$.

Practice Question

2. A particle accelerates according to $a(t) = 4t - 2$. Find the change in speed between $t=1$ and $t=4$.

Practice Question

3. If a particle's velocity is given by $v(t) = 5 \cos(t)$, calculate the distance traveled from $t=0$ to $t=2\pi$.

#Glossary

• Distance: The total path length traveled, regardless of direction.
• Displacement: The change in position of an object, having both magnitude and direction.
• Speed: The magnitude of velocity, always positive.
• Velocity: The rate of change of position, having both magnitude and direction.
• Integral: A mathematical operation that finds the area under a curve, often used to compute accumulated quantities such as distance.

#Summary and Key Takeaways

• Distance and displacement are distinct concepts; distance is a scalar while displacement is a vector.
• Speed is the magnitude of velocity and is always positive.
• Use integration to find the total distance traveled by considering the absolute value of the velocity function.
• Pay attention to whether a problem requires calculations with speed or velocity, especially when velocities change sign.
• Practice with various examples to solidify understanding and application of these concepts.

By mastering these concepts and practicing integration techniques, you will be well-prepared for related questions in your exams.