Question

A particle is moving along the $x-$ axis with its coordinate with the time ${}^{\prime }{t}^{\prime }$ given be $x\left(t\right)=10+8t-3t^2$. Another particle is moving the $y-$ axis with its coordinate as a function of time given by $y\left(t\right)=5-8t^3$. At $t=1s$, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$. Then $v$ (in $m/s)$ is

Answer 1

1. Find speed of the first particle at $t=1s$.

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

Given, the position of the first particle,

$x\left(t\right)=10+8t-3t^2$

${\overrightarrow{r}}_1=(10+8t=3t^2)\hat{l}$

So, its velocity at any time $t$,

${\overrightarrow{v}}_1=\frac{d{\overrightarrow{r}}_1}{dt}$

${\overrightarrow{v}}_1=(8-6t)\hat{l}$

So, velocity of first particle at $t=1s$ ,

${\overrightarrow{v}}_1=2\hat{l}$ $m/s$

2. Find speed of the second particle at $t=1s$.

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

Given, the position of the second particle,

$y\left(t\right)=5-8t^3$

${\overrightarrow{r}}_2=(5-8t^3)\hat{j}$

So, its velocity at any time $t$,

${\overrightarrow{v}}_2=\frac{d{\overrightarrow{r}}_2}{dt}$

${\overrightarrow{v}}_2=-24t^2\hat{j}$

So, velocity of second particle at $t=1s$,

${\overrightarrow{v}}_2=-24\hat{j}m/s$

3. Find $v$ .

Given, the speed of the second particle in the frame of the first particle $=\sqrt{v}$

Velocity of the second particle with respect to first,


${\overrightarrow{v}}_{2/1}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1$

${\overrightarrow{v}}_{2/1}=-24\hat{j}-2\hat{i}$

Velocity of the second particle with respect to first,

$\left|{\overrightarrow{v}}_{2/1}\right|=\sqrt{(2{)}^2+(24{)}^2}$

$\left|{\overrightarrow{v}}_{2/1}\right|=\sqrt{580}m/s$

$v=580m/s$

Final answer : $580$