Question

A particle starts from the origin at 𝑡=0 𝑠 with a velocity of $10.0\hat{j}m/s$ and moves in the x-y plane with constant acceleration of $8.0\hat{i}+2.0\hat{j}{\text{ms}}^{-2}$.

A) At what time is the x - coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

B) What is the speed of the particle at the time ?

Answer 1

A) Step 1 : - Find $\overrightarrow{v}\left(t\right)$ from relation $\overrightarrow{a}\left(t\right)=\frac{d\overrightarrow{v}\left(t\right)}{dt}$
$\int d\overrightarrow{v}\left(t\right)=\int \overrightarrow{a}\left(t\right)dt$
Given $\overrightarrow{a}\left(t\right)=8\hat{i}+2\hat{j}$
${\int }_{10\hat{j}}^{v}d\overrightarrow{v}\left(t\right)={\int }_0^t(8\hat{i}+2\hat{j})dt$
$\overrightarrow{v}\left(t\right)-10\hat{j}=8t\hat{i}+2t\hat{j}$
$\overrightarrow{v}\left(t\right)=8t\hat{i}+2t\hat{j}+10\hat{j}$
$\overrightarrow{v}\left(t\right)=8t\hat{i}+(2t+10)\hat{j}.......\left(1\right)$
Step 2 : - Find $\overrightarrow{r}\left(t\right)$ from relation $\overrightarrow{v}\left(t\right)=\frac{d\overrightarrow{r}\left(t\right)}{dt}$
As $d\overrightarrow{r}\left(t\right)=\overrightarrow{v}\left(t\right)dt$
${\int }_0^{r}d\overrightarrow{r}\left(t\right)={\int }_0^t\overrightarrow{v}\left(t\right)dt$
$\overrightarrow{r}\left(t\right)={\int }_0^t[8t\hat{i}+(2t+10\left)\hat{j}\right]dt$
$\overrightarrow{r}\left(t\right)=\frac{8t^2}{2}\hat{i}+(\frac{2t^2}{2}+10t)\hat{j}$
We can write $\overrightarrow{r}\left(t\right)=x\left(t\right)\hat{i}+y\left(t\right)\hat{j}=4t^2\hat{i}+(t^2+10)\hat{j}$
From comparsion: -
$x\left(t\right)=4t^2$
Putting $x\left(t\right)=16m$
$16=4t^2$
So, t = 2 s is the time when particle is at x = 16 m.
At t = 2s the y-coordinate of particle,
$y\left(t\right)=t^2+10=(2{)}^2+10=14m$

Final Answer: The particle at x = 16 m is at t = 2s and at t = 2s the y-coordinate of particle is 14 m.

B) Step 1 : Find the magnitude of velocity at $t=2\text{sec}$.
From equation (1)
$\overrightarrow{v}\left(t\right)=8t\hat{i}+(2t+10)\hat{j}$
Velocity at t = 2 s,
${\overrightarrow{v}}_{t=2s}=8\left(2\right)\hat{i}+\left(2\right(2)+10)\hat{j}$
${\overrightarrow{v}}_{t=2s}=16\hat{i}+14\hat{j}$
speed of particle : -
$|\overrightarrow{v}{|}_{t=2s}=\sqrt{(16{)}^2+(14{)}^2}$
$=\sqrt{256+196}$
$=\sqrt{452}$
$=21.26m/s$
Final Answer: Speed of the particle is 21.26 m/s.