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Explain why it is not possible to have an average velocity with a large magnitude but a small average speed.

Short Answer

Expert verified
Average speed cannot be smaller than the magnitude of average velocity, as speed considers the whole path, while velocity considers displacement.

Step by step solution

01

Define Average Velocity and Speed

Average velocity is defined as the total displacement divided by the total time taken, represented by the formula \( v_{avg} = \frac{\Delta x}{\Delta t} \). Average speed, on the other hand, is the total distance traveled divided by the total time taken, given by \( s_{avg} = \frac{d}{\Delta t} \). Displacement is a vector quantity considering direction, while distance is a scalar considering only magnitude.
02

Evaluate Magnitudes

Average velocity can have a large magnitude if there is a significant displacement in a short amount of time. However, it relies entirely on the net change in position. On the contrary, average speed measures the entire path taken, which is always equal to or greater than the displacement.
03

Compare Concepts

Since average speed accounts for the total path length, it can never be less than the magnitude of the average velocity. The magnitude of displacement cannot exceed the distance traveled. Therefore, for average velocity to have a large magnitude, the object must have a large displacement, implying a large path traveled, thus a large average speed.
04

Conclusion

Having a large average velocity means the displacement over time is large, which implies the total distance traveled is at least as large as the displacement. Consequently, it is not possible to have a large average velocity magnitude and a small average speed, because average speed will be equal to or larger than the magnitude of average velocity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Speed
Average speed might sound complicated, but it's actually quite simple! Imagine you're driving a car. Let's say you're covering a total distance over a period of time. Your average speed is calculated by dividing that total distance by the total time it took to travel that distance.
The important thing to remember is that average speed considers the whole path you traveled, without worrying about the direction.
- Formula: \( s_{avg} = \frac{d}{\Delta t} \)- It’s a scalar quantity, meaning it only has magnitude, and not direction. - It tells you how fast you're moving on average, throughout your journey. So, average speed is great for understanding how quickly you're moving in general, regardless of any twists or turns along the way.
Displacement
Displacement is a little bit different from distance. It’s all about where you end up relative to where you started. Think of displacement as a straight line drawn from your starting point to your ending point.
Displacement is a vector quantity, and here's what makes it special:- It considers both magnitude and direction.- Formula: It is often represented by \( \Delta x \) (change in position).- It’s the shortest path between the initial and final points. Imagine you’re walking north five steps, then turn east and walk another five steps. Your total distance is ten steps, but your displacement considers only the straight-line distance from start to finish, in a specific direction. This is why it can influence things like average velocity strongly.
Distance Traveled
Distance traveled is all about counting every single step along your journey. It measures the whole path you took, from start to finish. Think of it as a journey log that tracks all your movements regardless of direction.
What makes distance traveled simple yet essential? - It is always equal to or greater than the displacement. - Unlike displacement, it's a scalar quantity so it doesn’t consider direction, only how much ground you covered. So if you wandered around in circles before reaching your destination, distance traveled adds up everything, whereas displacement only sees the result.
Vector Quantity
Vector quantities are everywhere, especially when dealing with motion and physics. Any quantity that has both magnitude and direction is a vector. Displacement is a great example, while distance, lacking direction, is not.
Understanding vector quantity means understanding a few key points: - Magnitude tells you "how much" (e.g., how far, how fast). - Direction tells you "where" (e.g., north, east). - They are represented with arrows where length indicates magnitude and the arrowhead shows direction. Vectors are crucial in physics because they help us precisely describe the motion of objects. When you're working on problems involving vectors, you're getting a complete picture of what's happening, more than just with numbers alone.

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Most popular questions from this chapter

Ask you to verify parts of Theorem \(11.2 .4 .\) In each let \(f(t)=t^{3}, \vec{r}(t)=\left\langle t^{2}, t-1,1\right\rangle\) and \(\vec{s}(t)=\) \(\left\langle\sin t, e^{t}, t\right\rangle .\) Compute the various derivatives as indicated. Simplify \(\vec{r}(t) \cdot \vec{s}(t),\) then find its derivative; show this is the same as \(\vec{r}^{\prime}(t) \cdot \vec{s}(t)+\vec{r}(t) \cdot \vec{s}^{\prime}(t)\)

A curve \(C\) is described along with 2 points on \(C\). (a) Using a sketch, determine at which of these points the curvature is greater. (b) Find the curvature \(\kappa\) of \(C\), and evaluate \(\kappa\) at each of the 2 given points. \(C\) is defined by \(\vec{r}(t)=\left\langle t^{2}-1, t^{3}-t\right\rangle ;\) points given at \(t=0\) and \(t=5\).

A position function \(\vec{r}(t)\) of an object is given. Find the speed of the object in terms of \(t,\) and find where the speed is minimized/maximized on the indicated interval. $$ \vec{r}(t)=\left\langle t^{2}, t^{2}-t^{3}\right\rangle \text { on }[-1,1] $$

A position function \(\vec{r}(t)\) is given along with its unit tangent vector \(\vec{T}(t)\) evaluated at \(t=a\), for some value of \(a\) (a) Confirm that \(\vec{T}(a)\) is as stated. (b) Using a graph of \(\vec{r}(t)\) and Theorem 11.4.1, find \(\vec{N}(a)\). $$ \vec{r}(t)=\left\langle\cos ^{3} t, \sin ^{3} t\right\rangle ; \quad \vec{T}(\pi / 4)=\left\langle-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right\rangle $$

Position functions \(\vec{r}_{1}(t)\) and \(\vec{r}_{2}(s)\) for two objects are given that follow the same path on the respective intervals. (a) Show that the positions are the same at the indicated \(t_{0}\) and \(s_{0}\) values; i.e., show \(\vec{r}_{1}\left(t_{0}\right)=\vec{r}_{2}\left(s_{0}\right) .\) (b) Find the velocity, speed and acceleration of the two objects at \(t_{0}\) and \(s_{0},\) respectively. $$ \begin{array}{l} \vec{r}_{1}(t)=\left\langle t, t^{2}\right\rangle \text { on }[0,1] ; t_{0}=1 \\\ \vec{r}_{2}(s)=\left\langle s^{2}, s^{4}\right\rangle \text { on }[0,1] ; s_{0}=1 \end{array} $$

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