find the velocity, acceleration, and speed of a particle with the given position function.\nr(t) = 5 cos(t)i…

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)} )