Question

find the velocity, acceleration, and speed of a particle with the given position function.
r(t) = 5 cos(t)i + 4 sin(t)j
v(t) =
a(t) =
|v(t)| =

Explanation:

Step1: Find the velocity vector \( \mathbf{v}(t) \)

Differentiate the position function \( \mathbf{r}(t)=5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} \) with respect to \( t \).
Using the derivative rules \( \frac{d}{dt}(\cos(t))=-\sin(t) \) and \( \frac{d}{dt}(\sin(t))=\cos(t) \), we have:
\( \mathbf{v}(t)=\mathbf{r}'(t)=- 5\sin(t)\mathbf{i}+4\cos(t)\mathbf{j} \)

Step2: Find the acceleration vector \( \mathbf{a}(t) \)

Differentiate the velocity vector \( \mathbf{v}(t) \) with respect to \( t \).
Using the derivative rules \( \frac{d}{dt}(\sin(t))=\cos(t) \) and \( \frac{d}{dt}(\cos(t))=-\sin(t) \), we get:
\( \mathbf{a}(t)=\mathbf{v}'(t)=-5\cos(t)\mathbf{i}-4\sin(t)\mathbf{j} \)

Step3: Find the speed \( |\mathbf{v}(t)| \)

Use the formula for the magnitude of a vector \( \mathbf{v}(t) = x\mathbf{i}+y\mathbf{j} \), \( |\mathbf{v}(t)|=\sqrt{x^{2}+y^{2}} \).
Here \( x = - 5\sin(t) \) and \( y = 4\cos(t) \), so \( |\mathbf{v}(t)|=\sqrt{(-5\sin(t))^{2}+(4\cos(t))^{2}}=\sqrt{25\sin^{2}(t)+16\cos^{2}(t)}=\sqrt{16(\sin^{2}(t)+\cos^{2}(t)) + 9\sin^{2}(t)} \).
Since \( \sin^{2}(t)+\cos^{2}(t) = 1 \), we have \( |\mathbf{v}(t)|=\sqrt{16 + 9\sin^{2}(t)} \)

Answer:

\( \mathbf{v}(t)=-5\sin(t)\mathbf{i}+4\cos(t)\mathbf{j} \)
\( \mathbf{a}(t)=-5\cos(t)\mathbf{i}-4\sin(t)\mathbf{j} \)
\( |\mathbf{v}(t)|=\sqrt{25\sin^{2}(t)+16\cos^{2}(t)} \)