A positon vector of a particle at time $t$ is $R= Cos(t-1) i + Sinh(t-1) j + st^2 k$. If at $t = 1$, the acceleration of the particle be perpendicular to its position vector, find $s$.
I know that $dR/dt = velocity$ and $d^2R / dt^2 = acceleration$. How do I use the fact that R is perpendicular to acceleration to obtain the solution
Does it mean that at $t=1$, $d^2R/dt^2 = 0$.
I tried to proceed like so .
$R.R = Cos^2 (t-1) + Sinh^2(t-1) +s^2t^4$
$R= sqrt[(Cos^2 (t-1) + Sinh^2(t-1) + s^2t^4)]$
$2R dR/dt = ...$
$2 dR/dt + 2R d^2R/dt^2 = ...$
and then I get stuck.
Since acceleration is normal to position vector I have also thought weather (del $phi\ dR =0$) could be applied. but then I ask how?.
Hint: two vectors $\,\mathbf{u},\mathbf{v}\,$ are perpendicular iff their dot product $\,\langle \mathbf{u}, \mathbf{v} \rangle\,$ is zero. In this case:
$$ \begin{align} \begin{cases} R &= \cos(t-1) \,\mathbf{i} + \sinh(t-1) \,\mathbf{j} + st^2 \,\mathbf{k} \\ \ddot{R} = \dfrac{d^2 R}{d t^2} &= -\cos(t-1) \,\mathbf{i} + \sinh(t-1) \,\mathbf{j} + 2s \,\mathbf{k} \end{cases} \implies \begin{cases} R(1) = 1 \cdot \mathbf{i} + 0 \cdot\mathbf{j} + s \cdot \mathbf{k} \\[7px] \ddot{R}(1) = -1 \cdot \mathbf{i} + 0 \cdot\mathbf{j} + 2s \cdot \mathbf{k} \end{cases} \end{align} $$
The condition $\,\left\langle R(1), \,\ddot{R}(1)\right\rangle = 0\,$ then gives a simple equation to solve for $\,s\,$.