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Chapter 12: Problem 8

Consider the following position functions. a. Find the velocity and speed of the object. b. Find the acceleration of the object. $$\mathbf{r}(t)=\left\langle\frac{5}{2} t^{2}+3,6 t^{2}+10\right\rangle, \text { for } t \geq 0$$

Short Answer

Expert verified
Answer: The velocity function of the object is \(\mathbf{v}(t) = \left\langle 5t, 12t \right\rangle\), the speed is \(13t\), and the acceleration function is \(\mathbf{a}(t) = \left\langle 5, 12 \right\rangle\).

Step by step solution

01

Find the velocity function

To find the velocity function, \(\mathbf{v}(t)\), we need to find the first derivative of the position function, \(\mathbf{r}(t)\). So we have: $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}$$ Now we differentiate each component of the position function: $$\frac{d}{dt}(\frac{5}{2} t^2 + 3) = 5t$$ $$\frac{d}{dt}(6t^2 + 10) = 12t$$ Therefore, the velocity function is: $$\mathbf{v}(t) = \left\langle 5t, 12t \right\rangle$$
02

Find the speed of the object

Speed is the magnitude of the velocity vector, and can be found using the formula: $$\text{Speed} = |\mathbf{v}(t)| = \sqrt{(5t)^2 + (12t)^2}$$ Now, simplify the expression: $$\text{Speed} = \sqrt{25t^2 + 144t^2} = \sqrt{169t^2} = 13t$$ So the speed of the object is \(13t\).
03

Find the acceleration function

To find the acceleration function, \(\mathbf{a}(t)\), we need to find the second derivative of the position function, or the first derivative of the velocity function, \(\mathbf{v}(t)\). So we have: $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}$$ Now we differentiate each component of the velocity function: $$\frac{d}{dt}(5t) = 5$$ $$\frac{d}{dt}(12t) = 12$$ Therefore, the acceleration function is: $$\mathbf{a}(t) = \left\langle 5, 12 \right\rangle$$ So the velocity function is \(\mathbf{v}(t) = \left\langle 5t, 12t \right\rangle\), the speed of the object is \(13t\), and the acceleration function is \(\mathbf{a}(t) = \left\langle 5, 12 \right\rangle\).

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).

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