0
$\begingroup$

The position vector of moving point at time $t$ is $\mathbf{r}(t) = \left(\sin(t),\cos (2t),t^2+2t\right)$. Find components of acceleration $\mathbf{a}$ in directions parallel to velocity vector $\mathbf{v}$ and perpendicular to plane of $\mathbf{r}$ and $\mathbf{v}$ at $t=0.$

I know the acceleration vector is the second derivative of the $\mathbf{r}$ vector, but how to proceed?

$\endgroup$
3
  • $\begingroup$ By "... perpendicular to plane $\mathbf{r}$ and $\mathbf{v}$ ..." is the problem asking for the component of $\mathbf{a}$ that is in the plane containing $\mathbf{r}$ and $\mathbf{v}?$ $\endgroup$ – Adrian Keister Jun 26 '19 at 18:09
  • $\begingroup$ Aren't you able to compute this second derivative ? Please show us. $\endgroup$ – Yves Daoust Jun 26 '19 at 18:36
  • $\begingroup$ i have difficulty in understanding point of question $\endgroup$ – J. Deff Jun 27 '19 at 5:20
3
$\begingroup$

Given $$ \vec r = (\sin t, \cos 2t, t^2 + 2t)$$ $\Rightarrow \vec v = ( \cos t , -2 \sin 2t ,2t + 2)$ and $ \vec a = ( -\sin t , -4 \cos 2t , 2)$ which at $t = 0$ are $(1,0,0)$ and $( 0,-4,2)$ respectively.

Now,

  1. Projection of $\vec a$ on $\vec v$ is $\cfrac{ \vec a \cdot \vec v}{|v|} $ and the component parallel to $\vec v$ is $\cfrac{ \vec a \cdot \vec v}{|v|} \hat v$

  2. This part can be proceeded in the exact same way as the first part, you have replace $ \vec v$ with the normal vector of the plane (as that is the vector parallel to which you have to obtain the components) which here is $ \vec v \times \vec r$

$\endgroup$
12
  • 1
    $\begingroup$ I think you mean $\vec{v}\times\vec{r},$ not $\vec{v}\times\vec{a}.$ $\endgroup$ – Adrian Keister Jun 26 '19 at 18:18
  • $\begingroup$ can anyone explain what is going on $\endgroup$ – J. Deff Jun 26 '19 at 18:19
  • $\begingroup$ Which part @J.Deff ? $\endgroup$ – XRFXLP Jun 26 '19 at 18:19
  • $\begingroup$ @J.Deff Ajay is basically determining a unit vector in the required direction. Then finding the projection of the acceleration vector onto that unit vector. The second part involves finding a unit normal vector to the plane, which can be done by taking the cross product between two vectors lying on the plane, then dividing by its own magnitude. $\endgroup$ – Deepak Jun 26 '19 at 18:21
  • $\begingroup$ I have edited the part in which I thought you had the problem. $\endgroup$ – XRFXLP Jun 26 '19 at 18:22
0
$\begingroup$

$$\vec{r}(t)=(\sin t,\cos2t,t^2+2t)\Rightarrow \vec{r}(0)=(0,1,0)$$ $$\vec{v}(t)=(\cos t,-2\sin2t,2t+2)\Rightarrow \vec{v}(0)=(1,0,2)$$ $$\vec{a}(t)=(-\sin t,-4\cos2t,2)\Rightarrow \vec{a}(0)=(0,-4,2)$$ tangential acceleration is $$a_T(t)=\vec{a}(t)\cdot\vec{T}(t)=\vec{a}(t)\cdot\dfrac{\vec{r'}(t)}{|\vec{r'}(t)|}=\vec{a}(t)\cdot\dfrac{\vec{v}(t)}{|\vec{v}(t)|}$$ then $$a_T(0)=(0,-4,2)\cdot(\dfrac{1}{\sqrt{5}},0,\dfrac{2}{\sqrt{5}})=\dfrac{4}{\sqrt{5}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.