The position of a squirrel running in a park is given by r⃗=[(0.280 m/s)t+(0.0360 m/s2)t2]i^+(0.0190 m/s3)t3j^\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} r=[(0.280 m/s)t+(0.0360 m/s2)t2]i^+(0.0190 m/s3)t3j^. (a) What are vx(t)v_{x}\left(t\right) and vy(t)v_{y}\left(t\right), the xx-and yy-components of the velocity of the squirrel, as functions of time?

Ch 03: Motion in Two or Three Dimensions

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 03: Motion in Two or Three Dimensions

Problema 4a

Capitolo 3, Problema 4a

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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To find the velocity components, we need to differentiate the position vector with respect to time. The position vector is given as r(t) = [(0.280 m/s)t + (0.0360 m/s^2)t^2]î + (0.0190 m/s^3)t^3ĵ.

The x-component of the position is x(t) = (0.280 m/s)t + (0.0360 m/s^2)t^2. To find the x-component of the velocity, differentiate x(t) with respect to time t: υx(t) = d/dt[(0.280 m/s)t + (0.0360 m/s^2)t^2].

Apply the power rule of differentiation: d/dt[at^n] = n*at^(n-1). For the term (0.280 m/s)t, the derivative is 0.280 m/s. For the term (0.0360 m/s^2)t^2, the derivative is 2*(0.0360 m/s^2)t = 0.0720 m/s^2 * t.

Thus, the x-component of the velocity is υx(t) = 0.280 m/s + 0.0720 m/s^2 * t.

Now, for the y-component of the position, y(t) = (0.0190 m/s^3)t^3. Differentiate y(t) with respect to time t to find the y-component of the velocity: υy(t) = d/dt[(0.0190 m/s^3)t^3]. Using the power rule, the derivative is 3*(0.0190 m/s^3)t^2 = 0.0570 m/s^3 * t^2. Therefore, the y-component of the velocity is υy(t) = 0.0570 m/s^3 * t^2.

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Differentiation

Differentiation is a mathematical process used to find the rate at which a quantity changes. In physics, it is often used to determine velocity from a position function. By differentiating the position function with respect to time, we can find the velocity components, which are the derivatives of the position components.

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Vector Components

Vector components are the projections of a vector along the axes of a coordinate system. In this problem, the position vector r is expressed in terms of its x-component (î) and y-component (ĵ). Understanding how to separate these components is crucial for calculating the velocity in each direction independently.

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Polynomial Functions

Polynomial functions consist of terms with variables raised to integer powers. The position function given is a polynomial in terms of time, t, with coefficients indicating the rate of change. Recognizing the structure of polynomial functions helps in applying differentiation rules to find the velocity components as functions of time.

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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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