Question

A particle is moving in $x-y$ plane. At an instant 𝑡, the co-ordinates of a particle are $x=\frac{t^3}{3},y=2t^2$. At $t=3$, the angle between position vector and velocity vector of particle is ${\cos }^{-1}\left(\frac{x}{y\sqrt{z}}\right)$. Find $(z-x-y)$.

Answer 1

Position vector is given by
$\overrightarrow{r}=x\hat{i}+y\hat{j}$
Given $x=\frac{t^3}{3},y=2t^2$
$\overrightarrow{r}=\frac{t^3}{3}\hat{i}+2t^2\hat{j}$
At $t=3s$
$\overrightarrow{r}=\frac{3^3}{3}\hat{i}+2\times 3^2\hat{j}$
$=9\hat{i}+18\hat{j}$

$\overrightarrow{v}=\frac{dr}{dt}=\frac{3t^2}{3}\hat{i}+4t\hat{j}$
$\overrightarrow{v}=t^2\hat{i}+4t\hat{j}$

At $t=3s$
$\overrightarrow{v}=3^2\hat{i}+4\times 3\hat{j}$
$=9\hat{i}+12\hat{j}$

$\cos \theta =\frac{\overrightarrow{r}.\overrightarrow{v}}{\left|\overrightarrow{r}\overrightarrow{v}\right|}$
$\cos \theta =\frac{(9\times 9)+(18\times 12)}{\sqrt{9^2+{18}^2}.\sqrt{9^2+{12}^2}}$


$\cos \theta =\frac{81+216}{\sqrt{405}\sqrt{225}}$=$\frac{297}{\sqrt{405}\sqrt{225}}$

$\theta ={\cos }^{-1}\frac{297}{15\sqrt{405}}$

Given,

$\theta ={\cos }^{-1}\left(\frac{x}{y\sqrt{z}}\right)$

By comparing, we get
x=297, y=15, z=405
Hence, z-x-y=405-297-15=93.