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Consider the following trajectories of moving objects. Find the tangential and normal components of the acceleration. $$\mathbf{r}(t)=\left\langle t^{3}, t^{2}\right\rangle$$

Short Answer

Expert verified
Answer: The tangential component of acceleration, \(a_{T}(t)\), is given by \(\frac{12t^{3}+4t^{2}}{\sqrt{9t^{4} + 4t^{2}}}\), and the normal component of acceleration, \(a_{N}(t)\), is given by \(\frac{2}{\sqrt{9t^{4} + 4t^{2}}}\).

Step by step solution

01

Find the first and second derivatives of the position vector

Differentiate the position vector \(\mathbf{r}(t)\) with respect to time \(t\), to find the velocity vector, \(\mathbf{v}(t)\), and then differentiate \(\mathbf{v}(t)\) to find the acceleration vector, \(\mathbf{a}(t)\). $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \left\langle 3t^{2}, 2t\right\rangle$$ $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \left\langle 6t, 2\right\rangle$$
02

Calculate the magnitude of the velocity vector

Determine the magnitude of \(\mathbf{v}(t)\) using the formula: $$||\mathbf{v}(t)|| = \sqrt{(3t^{2})^{2} + (2t)^{2}} = \sqrt{9t^{4} + 4t^{2}}$$
03

Determine the unit tangent vector

Divide the velocity vector, \(\mathbf{v}(t)\), by its magnitude to find the unit tangent vector, \(\mathbf{T}(t)\). $$\mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} = \frac{\left\langle 3t^{2}, 2t\right\rangle}{\sqrt{9t^{4} + 4t^{2}}}$$
04

Calculate the tangential component of the acceleration

Find the tangential component of the acceleration, \(a_{T}(t)\), using the dot product of the acceleration vector, \(\mathbf{a}(t)\), and the unit tangent vector, \(\mathbf{T}(t)\). $$a_{T}(t) = \mathbf{a}(t) \cdot \mathbf{T}(t) = \left\langle 6t, 2\right\rangle \cdot \frac{\left\langle 3t^{2}, 2t\right\rangle}{\sqrt{9t^{4} + 4t^{2}}} = \frac{12t^{3}+4t^{2}}{\sqrt{9t^{4} + 4t^{2}}}$$
05

Calculate the normal component of the acceleration

Use the formula \(a_{N}(t) = \sqrt{a(t)^{2} - a_{T}(t)^{2}}\) to find the normal component of the acceleration. $$a_{N}(t) = \sqrt{||\mathbf{a}(t)||^{2} - a_{T}(t)^{2}} = \sqrt{\left( (6t)^{2} + (2)^{2} \right) - \left( \frac{12t^{3}+4t^{2}}{\sqrt{9t^{4} + 4t^{2}}} \right)^{2}} = \frac{2}{\sqrt{9t^{4} + 4t^{2}}}$$
06

Present the tangential and normal components of acceleration

The tangential component of acceleration, \(a_{T}(t)\), is given by: $$a_{T}(t) = \frac{12t^{3}+4t^{2}}{\sqrt{9t^{4} + 4t^{2}}}$$ And the normal component of acceleration, \(a_{N}(t)\), is given by: $$a_{N}(t) = \frac{2}{\sqrt{9t^{4} + 4t^{2}}}$$

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Show that the two-dimensional trajectory $$x(t)=u_{0} t+x_{0}\( and \)y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0},\( for \)0 \leq t \leq T$$ of an object moving in a gravitational field is a segment of a parabola for some value of \(T>0 .\) Find \(T\) such that \(y(T)=0\)

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