A web page designer creates an animation in which a dot on a computer screen has position r→=[4.0cm⁡+(2.5 cm/s2)t2]i+(5.0cm/s⁡)tj\overrightarrow{r}=\left\lbrack4.0\operatorname{cm}+\left(2.5\text{ cm/s}^2\$$right)t^2$$\right\rbrack\mathbf{i}+\left(5.0\operatorname{\text{cm/s}}\right)t\mathbf{j}. Find the magnitude and direction of the dot's average velocity between t=0t = 0 and t=2.0st=2.0s.

Ch 03: Motion in Two or Three Dimensions

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

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Young & Freedman Calc 14th Edition

Ch 03: Motion in Two or Three Dimensions

Problem 3a

Chapter 3, Problem 3a

A web page designer creates an animation in which a dot on a computer screen has position $\vec{r} = [4.0 cm + (2.5 {cm/s}^{2}) t^{2}] i + (5.0 cm/s) t j$ . Find the magnitude and direction of the dot's average velocity between $t equals 0$ and $t equals 2.0 s$ .

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Identify the position vector $ \mathbf{r}(t) = [4.0\, \text{cm} + (2.5\, \text{cm/s}^2)t^2]\mathbf{i} + (5.0\, \text{cm/s})t\mathbf{j} $.

Calculate the position vector at $ t = 0 $ by substituting $ t = 0 $ into $ \mathbf{r}(t) $.

Calculate the position vector at $ t = 2.0 $ s by substituting $ t = 2.0 $ s into $ \mathbf{r}(t) $.

Determine the change in position vector $ \Delta \mathbf{r} = \mathbf{r}(2.0) - \mathbf{r}(0) $.

Calculate the average velocity $ \mathbf{v}_{\text{avg}} = \frac{\Delta \mathbf{r}}{\Delta t} $ and find its magnitude and direction using $ \text{magnitude} = \sqrt{v_{x}^2 + v_{y}^2} $ and $ \text{direction} = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Average Velocity

Average velocity is defined as the total displacement divided by the total time taken. It is a vector quantity, meaning it has both magnitude and direction. To find the average velocity, calculate the change in position over the given time interval and divide by the duration of that interval.

Vector Components

Vectors have components that describe their influence in different directions, typically represented by î, ĵ, and k̂ for the x, y, and z axes respectively. In this problem, the position vector r→ is given in terms of its î and ĵ components, which need to be analyzed separately to determine the overall displacement and velocity.

Magnitude and Direction of a Vector

The magnitude of a vector is its length, calculated using the Pythagorean theorem for its components. The direction is given by the angle it makes with a reference axis, often found using trigonometric functions like tangent. For the average velocity, compute the magnitude from its components and use inverse tangent to find the direction.

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Textbook Question

The position of a squirrel running in a park is given by $\vec{r}\) = $\left$[ (0.280~$\mathrm{m/s}$)t + (0.0360~$\mathrm{m/s^2}$)t^2 $\right$] $\hat{i}$ + (0.0190~$\mathrm{m/s^3}$)t^3 \(\hat{j}$ . At $t equals 5.00 s$ , how far is the squirrel from its initial position?

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Textbook Question

The position of a squirrel running in a park is given by $\vec{r}\) = $\left$[ (0.280~$\mathrm{m/s}$)t + (0.0360~$\mathrm{m/s^2}$)t^2 $\right$] $\hat{i}$ + (0.0190~$\mathrm{m/s^3}$)t^3 \(\hat{j}$ . (a) What are $v_{x} (t)$ and $v_{y} (t)$ , the $x$ -and $y$ -components of the velocity of the squirrel, as functions of time?

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Textbook Question

A dog running in an open field has components of velocity v_x = 2.6 m/s and v_y = −1.8 m/s at t1 = 10.0 s. For the time interval from t₁ = 10.0 s to t₂ = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s² and direction 31.0° measured from the +x–axis toward the +y–axis. At t₂ = 20.0 s, what are the x- and y-components of the dog's velocity?

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